Everything on Exam 1, from zero

"Either one but not both" is the event $(A\cap B^c) \cup (A^c \cap B)$ — two cells of the grid, and they are disjoint, so you add them. Fill the table:

Start with the given 0.1 in the top-left. The $B$ row must total 0.3, so the cell beside it is $0.3-0.1=0.2$. The $A$ column must total 0.5, so the cell below is $0.5-0.1=0.4$. The last cell is whatever makes everything sum to 1: $0.3$.

Now read off the two "exactly one" cells: $0.4 + 0.2$.

0.6

Let $A$ = "makes the first", $B$ = "makes the second".

(a) You want $\mathbb{P}(A\cap B)$. You are given only $\mathbb{P}(A)$ and $\mathbb{P}(B)$. Two marginals do not determine an intersection — you would need independence, and nothing says the shots are independent (a player who makes the first may be more relaxed on the second).

Cannot be determined

(b) "Miss both" is $A^c \cap B^c$, so $\mathbb{P}(A^c\cap B^c)=0.1$. Use the neither-nor identity:

$$\mathbb{P}(A\cup B) = 1 - \mathbb{P}(A^c\cap B^c) = 1 - 0.1 = 0.9$$

Then the additive law, rearranged:

$$\mathbb{P}(A\cap B) = \mathbb{P}(A)+\mathbb{P}(B)-\mathbb{P}(A\cup B) = 0.8 + 0.6 - 0.9$$

0.5

(c) Again $\mathbb{P}(A\cup B)=1-0.1=0.9$. But now split $A\cup B$ into two disjoint pieces — "exactly one" and "both":

$$\mathbb{P}(A\cup B) = \mathbb{P}(\text{exactly one}) + \mathbb{P}(\text{both}) \;\Rightarrow\; 0.9 = 0.3 + \mathbb{P}(\text{both})$$

0.6

Sanity check: $\mathbb{P}(B) = \mathbb{P}(A\cup B) - \mathbb{P}(A) + \mathbb{P}(A\cap B) = 0.9-0.8+0.6 = 0.7$, which is a legal probability. Good.

(d) Let $X$ = points scored, so $X \in \{0,1,2\}$. You are given $\mathbb{P}(X=2)=0.4$ and $\mathbb{P}(X=0)=0.2$, so $\mathbb{P}(X=1) = 1-0.4-0.2 = 0.4$. "Average number of points" is the expected value (module 5):

$$\mathbb{E}(X) = 0(0.2) + 1(0.4) + 2(0.4) = 1.2$$

1.2 points

(a) Straight additive law: $0.95 + 0.80 - 0.78$.

0.97

(b) You want $\mathbb{P}(A^c \mid B)$. Use the conditional definition (module 3) and get the numerator by subtraction — the $B$ row splits into "B and A" plus "B and not A":

$$\mathbb{P}(A^c\cap B) = \mathbb{P}(B) - \mathbb{P}(A\cap B) = 0.80 - 0.78 = 0.02$$

$$\mathbb{P}(A^c\mid B) = \frac{0.02}{0.80} = 0.025$$

0.025

Note $\mathbb{P}(A^c)=0.05$ but $\mathbb{P}(A^c\mid B)=0.025$. Knowing B fired makes A's failure less likely. The events are not independent.

$\Omega$ has $6\times 6 = 36$ equally likely ordered pairs.

(a) First $\in\{2,6\}$ and second $\in\{1,5\}$ — every combination:

(2,1), (2,5), (6,1), (6,5)

(b) First $\in \{2,6\}$ and sum $>7$. If the first is 2, you need the second $>5$, so only 6. If the first is 6, you need the second $>1$, so 2 through 6.

(2,6), (6,2), (6,3), (6,4), (6,5), (6,6)

(c) Take the four pairs from (a) and keep those with sum $>7$: sums are 3, 7, 7, 11. Note that $7$ is not $>7$ — the inequality is strict.

(6,5) only

(d) 4 favourable simple events out of 36, each equally likely:

$$\mathbb{P}(A\cap B) = \frac{4}{36} = \frac{1}{9}$$

1/9

Cross-check with independence: the two rolls genuinely are independent, and $\mathbb{P}(A)=2/6$, $\mathbb{P}(B)=2/6$, so $\mathbb{P}(A)\mathbb{P}(B) = 4/36$. Matches.

(a) De Morgan: $(A\cap B)^c = A^c \cup B^c$. In words, "not both" means "at least one of them failed."

(b) $(A\cup B)^c = A^c \cap B^c$. "Neither."

(c) Read the second condition first. $A\cap B = B$ says the overlap is all of $B$, which means every outcome in $B$ is also in $A$: $B \subset A$. But then $A \cup B$ adds nothing new to $A$, so $A\cup B = A$. And you were told $A\cup B=\Omega$.

$A = \Omega$

(a) $\Omega = \{H,\; TH,\; TTH,\; TTTH,\; \ldots\}$ — infinitely many simple events.

(b) Flip the event. "Four or more tosses needed" happens exactly when the first three tosses are all tails — nothing about the fourth matters. By independence:

$$\mathbb{P}(TTT) = \left(\tfrac12\right)^3 = \tfrac18$$

1/8

(c) "First head on flip 2, or 4, or 6, ..." — these are disjoint, so add:

$$\mathbb{P} = \left(\tfrac12\right)^2 + \left(\tfrac12\right)^4 + \left(\tfrac12\right)^6 + \cdots = \tfrac14\left[1 + \tfrac14 + \left(\tfrac14\right)^2+\cdots\right]$$

Apply the given identity with $\mathbb{P}(A)=\tfrac14$ and $\mathbb{P}(B)=\tfrac14$:

$$= \frac{1/4}{1-1/4} = \frac{1/4}{3/4}$$

1/3

Sanity check: the odd flips must then carry probability $2/3$. That is reasonable — flip 1 alone already has probability $1/2$.

Step 1 — name the events and translate. Let $N$ = "contains nitrates", $R$ = "turns red". The phrase "if it contains nitrates, it turns red 90% of the time" means the condition is $N$ and the outcome is $R$:

$$\mathbb{P}(R\mid N)=0.9 \qquad \mathbb{P}(R\mid N^c)=0.1 \qquad \mathbb{P}(N)=0.3$$

(a) You want an unconditional probability from conditionals — that is L.o.T.P. The partition is $\{N, N^c\}$, which is disjoint and exhaustive:

$$\mathbb{P}(R) = \mathbb{P}(R\mid N)\mathbb{P}(N) + \mathbb{P}(R\mid N^c)\mathbb{P}(N^c) = (0.9)(0.3) + (0.1)(0.7) = 0.27 + 0.07$$

0.34

(b) Now you want $\mathbb{P}(N\mid R)$ — the reverse of what you were given. Bayes:

$$\mathbb{P}(N\mid R) = \frac{\mathbb{P}(R\mid N)\mathbb{P}(N)}{\mathbb{P}(R)} = \frac{(0.9)(0.3)}{0.34} = \frac{0.27}{0.34}$$

27/34 ≈ 0.794

Read the tree: of the total 0.34 worth of red specimens, 0.27 came from the nitrate branch. The answer is just that share. Notice the test looks impressive (90% sensitive) but a red result still only makes nitrates about 79% likely — because 70% of specimens had no nitrates to begin with, and 10% of those turn red too.

Let $C$ = "current flows" and let $1, 2, 3$ denote "relay $i$ is closed", each with probability 0.9.

(a) Current flows if relay 1 or 2 or 3 works, so $C = 1\cup 2\cup 3$. Apply the additive law twice, nesting:

$$\mathbb{P}(1\cup 2) = \mathbb{P}(1)+\mathbb{P}(2)-\mathbb{P}(1\cap 2) = 0.9+0.9-0.9^2 = 0.99$$

$$\mathbb{P}(C) = \mathbb{P}\big((1\cup2)\cup 3\big) = 0.99 + 0.9 - (0.99)(0.9) = 0.999$$

0.999

(b) You want $\mathbb{P}(1\mid C)$. The key observation: if relay 1 is operational, current necessarily flows, so the event $1$ is a subset of $C$, which means $1\cap C = 1$.

$$\mathbb{P}(1\mid C) = \frac{\mathbb{P}(1\cap C)}{\mathbb{P}(C)} = \frac{\mathbb{P}(1)}{\mathbb{P}(C)} = \frac{0.9}{0.999}$$

≈ 0.901

Each bowl is chosen with probability $1/3$. "Without replacement" means the second draw's probability depends on the first — the denominator drops from 4 to 3.

(a) Compute $\mathbb{P}(WW\mid \text{bowl } i)$ for each bowl by chaining the two draws:

Bowl 1 gives zero because after taking its only white ball there are none left. Now L.o.T.P. over the three bowls:

$$\mathbb{P}(WW) = \tfrac13\left(0 + \tfrac16 + \tfrac12\right) = \tfrac13\cdot\tfrac23$$

2/9

(b) Let $D$ = "one white and one black." Careful: this can happen as $WB$ or $BW$, so add both orders.

$$\mathbb{P}(D) = \tfrac13\left(\tfrac12+\tfrac23+\tfrac12\right) = \tfrac13 \cdot \tfrac53 = \tfrac59$$

Bayes:

$$\mathbb{P}(\text{Bowl 1}\mid D) = \frac{\mathbb{P}(D\mid \text{B1})\,\mathbb{P}(\text{B1})}{\mathbb{P}(D)} = \frac{\tfrac12\cdot\tfrac13}{\tfrac59} = \frac{1/6}{5/9} = \frac16\cdot\frac95$$

3/10

Below the prior of $1/3$, which makes sense: Bowl 2 is the most likely producer of a mixed pair, so seeing one shifts belief away from Bowls 1 and 3.

Let $K$ = "knows", $G = K^c$ = "guesses", $R$ = "answers correctly". The unstated given: if she knows the answer she gets it right, so $\mathbb{P}(R\mid K)=1$.

$$\mathbb{P}(R) = \mathbb{P}(R\mid K)\mathbb{P}(K) + \mathbb{P}(R\mid G)\mathbb{P}(G) = (1)(0.8) + (0.5)(0.2) = 0.9$$

(a) $$\mathbb{P}(K\mid R) = \frac{(1)(0.8)}{0.9}$$

8/9 ≈ 0.889

(b) This one is not conditional — it asks for an intersection. Multiplicative law:

$$\mathbb{P}(G\cap R) = \mathbb{P}(R\mid G)\mathbb{P}(G) = (0.5)(0.2)$$

0.1

Watch the wording. "Both A and B" is an intersection; "A given B" is conditional. Part (b) is the second leaf of the tree, not a ratio.

The hard part is the timing, not the probability. Work out how many donors you can afford to test.

If you test $k$ donors before finding a match, that consumes $2k$ minutes, and then the transfusion needs 2 more. So you need

$$2k + 2 \le 10 \quad\Longrightarrow\quad k \le 4$$

You get four attempts. The victim is saved if at least one of the first four donors is a match. Flip it with the complement rule — "at least one match" is the opposite of "all four fail":

$$\mathbb{P}(\text{saved}) = 1 - \mathbb{P}(\text{first four all fail}) = 1 - (0.6)^4 = 1 - 0.1296$$

0.8704

The independence here is physical: whether one donor matches says nothing about the next. Note this is also a binomial "at least one" — see module 6.

Probability for a continuous variable is area, so integrate the density over the region you want:

$$\mathbb{P}(X>0.5) = \int_{0.5}^{1} 3x^2\,dx = \Big[x^3\Big]_{0.5}^{1} = 1 - (0.5)^3 = 1 - \tfrac18$$

7/8 = 0.875

Note the upper limit is 1, not $\infty$ — the density is zero outside $[0,1]$. Also note $f_X(1)=3$, a density well above 1, which is fine.

(i) Masses must sum to 1:

$$p_X(1) = 1 - 0.20 - 0.35 - 0.15 = 0.30$$

For $\mathbb{P}(1\le X<3)$, ask which masses lie in the range. The inequality includes 1 and excludes 3, so it is $x=1$ and $x=2$:

$$0.30 + 0.35 = 0.65$$

$p_X(1)=0.30$,   $\mathbb{P}(1\le X<3)=0.65$

Interpretation, in your professor's words: "the next time we run the experiment governed by this distribution, there is a 65% chance we observe an outcome of $X=1$ or $X=2$."

(ii) Densities integrate to 1:

$$\int_0^1 cx^2\,dx = c\left[\tfrac{x^3}{3}\right]_0^1 = \tfrac{c}{3} = 1 \;\Longrightarrow\; c = 3$$

c = 3

(a) $$\int_0^3 cx^2\,dx = c\left[\tfrac{x^3}{3}\right]_0^3 = c\cdot 9 = 1 \;\Longrightarrow\; c=\tfrac19$$

c = 1/9

(b) $$\mathbb{P}(X<1) = \int_0^1 \tfrac19 x^2\,dx = \tfrac19\cdot\tfrac13 = \tfrac{1}{27} \approx 0.037$$

1/27 ≈ 0.037

Interpretation: if we run this experiment once more, there is about a 3.7% chance the value we observe is below 1.

(c) First get the conditioning event's probability:

$$\mathbb{P}(X<2) = \int_0^2 \tfrac19 x^2\,dx = \tfrac19\cdot\tfrac83 = \tfrac{8}{27}$$

Since $\{X<1\}\subset\{X<2\}$, the intersection collapses:

$$\mathbb{P}(X<1\mid X<2) = \frac{1/27}{8/27} = \frac18$$

1/8 = 0.125

Interpretation: among the runs that land below 2, one in eight lands below 1. Knowing the value is small makes a very small value more plausible — the probability rose from 3.7% to 12.5%.

(a) First, what values can $X$ take? You stop as soon as you hold both colours, so the earliest is 2. The latest is 4 — if you draw $BBB$ you have exhausted the black balls, so the fourth draw is certainly red. So $X\in\{2,3,4\}$.

$X=2$: the first two draws differ.

$$\mathbb{P}(RB) + \mathbb{P}(BR) = \tfrac25\cdot\tfrac34 + \tfrac35\cdot\tfrac24 = \tfrac{6}{20}+\tfrac{6}{20} = \tfrac{12}{20}$$

$X=3$: the first two match, the third breaks the streak. Two sub-cases.

• $RR$ then black: $\mathbb{P}(RR) = \tfrac25\cdot\tfrac14 = \tfrac{2}{20}$. Only blacks remain, so the third is certainly black. Contributes $\tfrac{2}{20}$.
• $BB$ then red: $\mathbb{P}(BB) = \tfrac35\cdot\tfrac24 = \tfrac{6}{20}$. Now 2 red and 1 black remain, so red comes with probability $\tfrac23$. Contributes $\tfrac{6}{20}\cdot\tfrac23 = \tfrac{4}{20}$.

$$\mathbb{P}(X=3) = \tfrac{2}{20}+\tfrac{4}{20} = \tfrac{6}{20}$$

$X=4$: the only route is $BBB$ then a forced red.

$$\mathbb{P}(BBB) = \tfrac35\cdot\tfrac24\cdot\tfrac13 = \tfrac{6}{60} = \tfrac{2}{20}$$

Check: $0.6+0.3+0.1 = 1$. Always do this.

(b) $$\mathbb{E}(X) = 2(0.6)+3(0.3)+4(0.1) = 1.2+0.9+0.4 = 2.5$$

E(X) = 2.5 draws

(c) Use the shortcut formula (module 5):

$$\mathbb{E}(X^2) = 4(0.6)+9(0.3)+16(0.1) = 2.4+2.7+1.6 = 6.7$$

$$\mathbb{V}(X) = \mathbb{E}(X^2)-\big(\mathbb{E}(X)\big)^2 = 6.7 - 6.25 = 0.45$$

σ = √0.45 ≈ 0.671

The principle: total probability is still 1. Sum the discrete masses, integrate the continuous density, set the whole thing equal to 1.

(a) $$0.25 + \int_{-1}^{1} c\,dx + 0.25 = 1 \;\Longrightarrow\; 0.5 + 2c = 1 \;\Longrightarrow\; c = 0.25$$

c = 0.25

(b) Same split: sum $x\cdot p$ over the masses, integrate $x\cdot f$ over the continuous part.

$$\mathbb{E}(X) = (-3)(0.25) + \int_{-1}^{1}x(0.25)\,dx + (2)(0.25)$$

The integral is zero by symmetry — a flat density on an interval centred at 0:

$$\int_{-1}^1 0.25x\,dx = 0.25\left[\tfrac{x^2}{2}\right]_{-1}^{1} = 0.25\left(\tfrac12 - \tfrac12\right) = 0$$

$$\mathbb{E}(X) = -0.75 + 0 + 0.5$$

E(X) = −0.25

Three lines, every time.

$$\mathbb{E}(X) = 0(0.4)+1(0.4)+2(0.2) = 0.8$$

$$\mathbb{E}(X^2) = 0^2(0.4)+1^2(0.4)+2^2(0.2) = 0.4+0.8 = 1.2$$

$$\mathbb{V}(X) = 1.2 - (0.8)^2 = 1.2-0.64 = 0.56$$

σ = √0.56 ≈ 0.748

Notice the squares land on the $x$ values, not on the probabilities. $2^2(0.2) = 0.8$, not $2(0.2^2)$.

There are 13 ranks, each equally likely. Group them by payoff:

• 2 through 10 — 9 ranks — lose 5
• J, Q — 2 ranks — win 10
• K, A — 2 ranks — win 4

Define $X$ = dollars gained, and write its p.m.f.:

$$\mathbb{E}(X) = -5\left(\tfrac{9}{13}\right) + 10\left(\tfrac{2}{13}\right) + 4\left(\tfrac{2}{13}\right) = \frac{-45+20+8}{13}$$

−17/13 ≈ −1.31 dollars per play

The lesson is the modelling step, not the arithmetic: define a random variable for the quantity you actually care about (money), build its p.m.f. by grouping outcomes with the same value, then apply the definition.

$$\mathbb{E}(X) = \int_0^1 x\cdot\theta x^{\theta-1}\,dx = \theta\int_0^1 x^{\theta}\,dx$$

Pull the constant $\theta$ out, then combine the powers: $x \cdot x^{\theta-1} = x^{\theta}$. Now integrate by the power rule:

$$= \theta\left[\frac{x^{\theta+1}}{\theta+1}\right]_0^1 = \frac{\theta}{\theta+1}(1 - 0)$$

E(X) = θ/(θ+1)

Interpretation: if we sample a single datum from this distribution, on average its value will be $\theta/(\theta+1)$.

Sanity check the behaviour: as $\theta$ grows, $\theta/(\theta+1)\to 1$, meaning large $\theta$ piles the density up near $x=1$. As $\theta\to 0$, the mean goes to 0. That is exactly the "how does $\theta$ change the shape" question your study guide asks about.

(a) Two unknowns need two equations. Call them $a = p_X(0)$ and $b = p_X(2)$.

Equation 1 — masses sum to 1:

$$a + 0.3 + b + 0.3 = 1 \;\Longrightarrow\; a+b = 0.4$$

Equation 2 — the given expected value:

$$0(a) + 1(0.3) + 2(b) + 3(0.3) = 1.8 \;\Longrightarrow\; 1.2 + 2b = 1.8 \;\Longrightarrow\; b = 0.3$$

p(0) = 0.1,   p(2) = 0.3

(b) $$\mathbb{E}(X^2) = 0(0.1)+1(0.3)+4(0.3)+9(0.3) = 0.3+1.2+2.7 = 4.2$$

$$\mathbb{V}(X) = 4.2 - (1.8)^2 = 4.2-3.24 = 0.96$$

σ = √0.96 ≈ 0.980

(c) Use the transformation rules rather than rebuilding the p.m.f. of $Y$. Here $a=3$, $b=0$:

$$\mathbb{E}(Y) = 3\,\mathbb{E}(X) = 3(1.8) = 5.4 \qquad \mathbb{V}(Y) = 3^2\,\mathbb{V}(X) = 9(0.96) = 8.64$$

E(Y) = 5.4,   V(Y) = 8.64

The common error is $\mathbb{V}(Y)=3(0.96)$. The scale factor enters squared.

Flip the event. "At least one tail" fails only when every flip is a head.

$$\mathbb{P}(\ge 1\text{ tail}) = 1 - \mathbb{P}(\text{no tails}) = 1 - \mathbb{P}(\text{3 heads}) = 1 - (0.4)^3 = 1 - 0.064$$

0.936 = 117/125

Formally, let $X$ = number of tails with $X\sim\mathrm{Binom}(3, 0.6)$; then $\mathbb{P}(X\ge1)=1-\binom{3}{0}(0.6)^0(0.4)^3$. Same number, but the shortcut is faster.

First, identify the distribution. The number of rolls is fixed in advance, so this is binomial.

(a) $X$ = number of 6s, $X\sim\mathrm{Binom}(4, 1/6)$:

$$\mathbb{P}(X\ge1) = 1-\mathbb{P}(X=0) = 1 - \binom{4}{0}\left(\tfrac16\right)^0\left(1-\tfrac16\right)^{4} = 1-\left(\tfrac56\right)^4 \approx 0.518$$

(b) Two dice give 36 equally likely pairs, so a double-6 has probability $1/36$. $X\sim\mathrm{Binom}(24, 1/36)$:

$$\mathbb{P}(X\ge1) = 1-\left(1-\tfrac1{36}\right)^{24} \approx 0.491$$

No — 0.518 vs 0.491

A real historical bet turned on this narrow gap. The point for you: identifying $k$ and $p$ correctly is the whole problem.

The structure: A can win in exactly 3, 4, or 5 games. Those are disjoint events, so their probabilities add.

1. Wins in 3. A must win all three: $p^3$.
2. Wins in 4. A wins exactly 2 of the first 3, then wins game 4. The first part is binomial, the second is a single independent game: $$\binom{3}{2}p^2(1-p)^{1}\cdot p = 3p^3(1-p)$$
3. Wins in 5. A wins exactly 2 of the first 4, then wins game 5: $$\binom{4}{2}p^2(1-p)^{2}\cdot p = 6p^3(1-p)^2$$

Add and factor out $p^3$:

$$\mathbb{P} = p^3\left[1 + 3(1-p) + 6(1-p)^2\right] = p^3\left[1 + 3 - 3p + 6 - 12p + 6p^2\right]$$

P = p³(6p² − 15p + 10)

Check at $p=0.5$: $0.125(1.5-7.5+10)=0.125(4)=0.5$. Even teams split the series half the time. Check at $p=1$: $6-15+10=1$. Both correct.

Step 1 — convert the rate into $\lambda$.

$$\lambda = 10\,\Delta t$$

Step 2 — write the probability of zero.

$$\mathbb{P}(X=0) = e^{-\lambda} = e^{-10\Delta t}$$

Step 3 — impose the condition and solve.

$$e^{-10\Delta t}\ \ge\ \tfrac12 \;\Longrightarrow\; -10\Delta t \ \ge\ \ln\left(\tfrac12\right) \;\Longrightarrow\; 10\Delta t\ \le\ \ln 2$$

$$\Delta t \le \frac{\ln 2}{10}$$

Δt ≤ 0.0693 minutes

Watch the inequality: multiplying by $-1$ flips it. And $-\ln(1/2)=\ln 2$.

(a) "No apparent flaws" means $X=0$, and you are told that happens 4% of the time. So work backwards from the zero-count formula:

$$\mathbb{P}(X=0) = e^{-\lambda} = 0.04 \;\Longrightarrow\; -\lambda = \ln(0.04) \;\Longrightarrow\; \lambda = \ln(25)$$

λ = ln 25 ≈ 3.219

(b) "More than two" means $X\ge3$. Complement it — three terms is much less work than an infinite sum:

$$\mathbb{P}(X>2) = 1 - \left[p_X(0)+p_X(1)+p_X(2)\right] = 1 - e^{-\lambda}\left[1 + \lambda + \frac{\lambda^2}{2}\right]$$

With $e^{-\lambda}=0.04$ and $\lambda=3.219$:

$$= 1 - 0.04\left[1 + 3.219 + 5.181\right] = 1 - 0.04(9.400) = 1-0.376$$

≈ 0.624

Notice $e^{-\lambda}$ was handed to you as 0.04 — substituting it saves you from exponentiating anything.

(c) For a Poisson, $\mathbb{E}(X)=\mathbb{V}(X)=\lambda$:

E(X) ≈ 3.219 flaws,   σ = √3.219 ≈ 1.794

Here $\mu=0$ and $\sigma=1$, so "$|X|>1$" is exactly "outside $\mu\pm\sigma$".

(a) First, the per-draw stopping probability:

$$\mathbb{P}(|X|>1) = 1 - \mathbb{P}(-1\le X\le 1) = 1-0.683 = 0.317$$

You stop within two draws unless both draws fail to exceed the threshold. Draws are independent:

$$\mathbb{P}(\text{stop by draw 2}) = 1 - (0.683)^2 = 1 - 0.4665$$

≈ 0.5335

Equivalently $0.317 + (0.683)(0.317)$ — stop on the first, or miss then stop on the second. Same number.

(b) Turn it into a ratio of unconditional probabilities. Since $\{X>2\}\subset\{X>1\}$:

$$\mathbb{P}(X>2\mid X>1) = \frac{\mathbb{P}(X>2)}{\mathbb{P}(X>1)}$$

Get each tail by halving the leftover:

$$\mathbb{P}(X>1) = \frac{1-0.683}{2} = 0.1585 \qquad \mathbb{P}(X>2) = \frac{1-0.954}{2} = 0.023$$

$$\frac{0.023}{0.1585} \approx 0.145$$

≈ 0.145

$$\mathbb{P}(X<2) = \int_0^2 \tfrac12 e^{-x/2}\,dx = \Big[-e^{-x/2}\Big]_0^2 = -e^{-1} - (-e^{0}) = 1 - e^{-1}$$

1 − e⁻¹ ≈ 0.632

This is the general fact that $\mathbb{P}(X<\theta) = 1-e^{-1}$ for any exponential: the probability of landing below the mean is always about 63.2%. That is what the skew does.

(i) With $\theta=1$ the density is just $e^{-x}$:

$$\mathbb{P}(1\le X\le 2) = \int_1^2 e^{-x}\,dx = \Big[-e^{-x}\Big]_1^2 = e^{-1}-e^{-2} \approx 0.368 - 0.135$$

≈ 0.233

Interpretation: the next time we run this experiment, there is a 23.3% chance of observing a value between 1 and 2.

(ii) You are solving for the parameter now, not the probability. Use the c.d.f.:

$$1 - e^{-1/\theta} = \tfrac12 \;\Longrightarrow\; e^{-1/\theta} = \tfrac12 \;\Longrightarrow\; -\tfrac1\theta = \ln\left(\tfrac12\right) \;\Longrightarrow\; \tfrac1\theta = \ln 2$$

θ = 1/ln2 ≈ 1.443

This $\theta$ is the value whose median is 1. Since the median is always below the mean for an exponential, $\theta>1$ as expected.

1. The likelihood. Multiply $n$ copies of the density, one per data point:

$$\mathcal{L}(\theta\mid\mathbf{x}) = \prod_{i=1}^{n}\frac1\theta e^{-x_i/\theta} = \frac{1}{\theta^n}\,e^{-\sum_{i=1}^n x_i/\theta}$$

The $n$ copies of $1/\theta$ multiply into $\theta^{-n}$. The exponents add, because $e^a e^b = e^{a+b}$.

2. Take the log.

$$\ell(\theta\mid\mathbf{x}) = \ln\!\left(\theta^{-n}\right) + \ln\!\left(e^{-\sum x_i/\theta}\right) = -n\ln\theta - \frac{1}{\theta}\sum_{i=1}^{n}x_i$$

3. Differentiate. Treat $\sum x_i$ as a constant — the data are fixed.

$$\ell'(\theta\mid\mathbf{x}) = -\frac{n}{\theta} + \frac{1}{\theta^2}\sum_{i=1}^{n}x_i$$

The second term: $-\tfrac1\theta \sum x_i = -\theta^{-1}\sum x_i$, whose derivative is $+\theta^{-2}\sum x_i$.

4. Set to zero.

$$\frac{1}{\theta^2}\sum_{i=1}^n x_i = \frac{n}{\theta}$$

5. Solve. Multiply both sides by $\theta^2$ and divide by $n$:

$$\theta = \frac1n\sum_{i=1}^{n}x_i$$

θ̂_MLE = X̄

Capitalized, because once you regard it as a rule applied to random data it is itself a random variable. And it is deeply sensible: $\mathbb{E}(X)=\theta$ for an exponential, so estimating $\theta$ by the sample mean is exactly what intuition suggests.

1. $$\mathcal{L}(p\mid\mathbf{x}) = \prod_{i=1}^{n}p^{x_i}(1-p)^{1-x_i} = p^{\sum x_i}\,(1-p)^{\,n-\sum x_i}$$

Multiplying powers of the same base adds the exponents: $\sum x_i$ for the first, and $\sum(1-x_i) = n - \sum x_i$ for the second.

2. $$\ell = \left(\sum x_i\right)\ln p + \left(n - \sum x_i\right)\ln(1-p)$$

3. $$\ell'(p) = \frac{\sum x_i}{p} - \frac{n-\sum x_i}{1-p}$$

The minus sign in the second term comes from the chain rule on $\ln(1-p)$.

4 & 5. Set to zero and cross-multiply:

$$\frac{1-p}{p} = \frac{n-\sum x_i}{\sum x_i} \;\Longrightarrow\; \frac1p - 1 = \frac{n}{\sum x_i} - 1 \;\Longrightarrow\; p = \frac1n\sum_{i=1}^n x_i$$

p̂_MLE = X̄

Now the invariance property. No new derivation needed — just substitute:

(√p)̂_MLE = √X̄

(p²)̂_MLE = X̄²

With $n=1$ the likelihood is just the density itself: $\mathcal{L}(\theta\mid x) = \theta x^{\theta-1}$.

Take the log. Use $\ln(AB)=\ln A + \ln B$ and then $\ln(A^n)=n\ln A$:

$$\ell = \ln\theta + (\theta-1)\ln x$$

Differentiate with respect to $\theta$. Here $\ln x$ is a constant — the datum is fixed — and $(\theta-1)\ln x$ is linear in $\theta$:

$$\frac{d\ell}{d\theta} = \frac{1}{\theta} + \ln x$$

Set to zero and solve.

$$\frac1\theta = -\ln x \;\Longrightarrow\; \theta = \frac{-1}{\ln x}$$

θ̂_MLE = −1 / ln(x)

Sanity check the sign: $x\in[0,1]$, so $\ln x \le 0$, so $-1/\ln x > 0$. The estimate respects the constraint $\theta>0$, as it must.

(a) Multiply $n$ copies and pull out the $\theta$s:

$$\mathcal{L}(\theta\mid\mathbf{x}) = \prod_{i=1}^{n}\theta(1-x_i)^{\theta-1} = \theta^n\prod_{i=1}^{n}(1-x_i)^{\theta-1} = \theta^n\left[\prod_{i=1}^{n}(1-x_i)\right]^{\theta-1}$$

The last step works because every factor carries the same exponent $\theta-1$.

(b) $$\ell = \ln\left(\theta^n\right) + (\theta-1)\ln\left[\prod_{i=1}^n(1-x_i)\right] = n\ln\theta + (\theta-1)\sum_{i=1}^{n}\ln(1-x_i)$$

The log of a product becomes a sum of logs — this is the step that makes the derivative easy.

(c) Differentiate. The sum is a constant with respect to $\theta$:

$$\frac{d\ell}{d\theta} = \frac{n}{\theta} + \sum_{i=1}^{n}\ln(1-x_i) = 0$$

$$\frac{n}{\theta} = -\sum_{i=1}^n \ln(1-x_i)$$

θ̂_MLE = −n / ∑ ln(1 − x_i)

Again check the sign: each $x_i\in[0,1]$ so $1-x_i\le1$ so $\ln(1-x_i)\le 0$, making the sum negative and the estimate positive.

(d) Pure invariance. The mean is $g(\theta) = 1/(1+\theta)$, so:

mean̂_MLE = 1 / (1 + θ̂_MLE)

$$\mathbb{E}(\bar{X}) = \mathbb{E}\!\left(\frac1n\sum_{i=1}^{n}X_i\right) = \frac1n\,\mathbb{E}\!\left(\sum_{i=1}^n X_i\right) = \frac1n\sum_{i=1}^{n}\mathbb{E}(X_i)$$

Each $X_i$ is drawn from the same distribution (that is the "identically distributed" part of i.i.d.), so every $\mathbb{E}(X_i)=\mu$:

$$= \frac1n\sum_{i=1}^{n}\mu = \frac1n\cdot n\mu = \mu$$

$$B(\bar{X}) = \mathbb{E}(\bar{X}) - \mu = 0$$

Unbiased — and for every distribution, not just some

Note this derivation used only linearity of expectation, which needs no independence. Independence enters when you compute the variance — module 9.

Apply the definition directly. Here $\hat{\theta} = \max(x_1,\ldots,x_n)$ and $\mathbb{E}(\hat{\theta}) = \dfrac{n\theta}{n+1}$:

$$B(\hat{\theta}) = \mathbb{E}(\hat{\theta}) - \theta = \frac{n\theta}{n+1} - \theta = \frac{n\theta - \theta(n+1)}{n+1} = \frac{n\theta - n\theta - \theta}{n+1}$$

B(θ̂) = −θ/(n+1)

The check: as $n\to\infty$, the denominator grows without bound, so $B\to 0$. The estimator is biased for any finite sample but becomes unbiased in the limit — exactly the asymptotic behaviour lecture 4W describes for MLEs.

Expected value. Pull the constant $1/n$ out, then distribute the expectation over the sum:

$$\mathbb{E}(\bar{X}) = \mathbb{E}\!\left(\frac1n\sum_{i=1}^n X_i\right) = \frac1n\mathbb{E}\!\left(\sum_{i=1}^n X_i\right) = \frac1n\sum_{i=1}^{n}\mathbb{E}(X_i) = \frac1n\sum_{i=1}^n \mu = \frac1n \cdot n\mu$$

E(X̄) = μ

Your professor's note: this step is true for any set of $X_i$ — linearity of expectation needs no assumptions. And the conclusion holds for every distribution: the expected value of the sample mean is the true mean.

Variance. Now the constant comes out squared:

$$\mathbb{V}(\bar{X}) = \mathbb{V}\!\left(\frac1n\sum_{i=1}^n X_i\right) = \frac{1}{n^2}\mathbb{V}\!\left(\sum_{i=1}^n X_i\right) = \frac{1}{n^2}\sum_{i=1}^{n}\mathbb{V}(X_i) = \frac{1}{n^2}\sum_{i=1}^n \sigma^2 = \frac{n\sigma^2}{n^2}$$

V(X̄) = σ²/n

The middle step is only valid for independent data. This is the one place in the derivation where independence is load-bearing — your professor marks it explicitly.

Standard error.

SE(X̄) = σ/√n

What this means: more data means less uncertainty about the true mean. But the $\sqrt{n}$ is important — to halve your uncertainty you need four times the data.

Step 1 — identify $n$ and $k$. Each datum is the number of heads in $k=3$ flips, so each $X_i\sim\mathrm{Binom}(3,p)$. You have six data values, so $n=6$.

Step 2 — the sufficient statistic for $p$ is the total.

$$X_+ = \sum_{i=1}^{6}X_i = 2+2+1+2+3+1 = 11$$

Step 3 — read its sampling distribution off the table. $X_+\sim\mathrm{Binom}(nk, p)$ with $nk = 6\times 3 = 18$.

X₊ = 11,   and   X₊ ~ Binom(18, p)

The logic is physical: you flipped the coin 18 times in total and saw 11 heads. Whether those flips came in groups of three is irrelevant to what they tell you about $p$ — which is exactly what "sufficient" means. And $p$ stays as a symbol; you are describing the distribution, not estimating the parameter.

Step 1 — check you may use the CLT. The distribution is unknown, so the table is no help. But $n=100\ge30$, the data are i.i.d., and the question is about the mean. CLT applies.

Step 2 — write the approximate sampling distribution.

$$\bar{X}\;\dot\sim\;\mathcal{N}\!\left(20,\;\frac{4}{100}\right) \qquad \mathrm{SE}(\bar X) = \frac{\sigma}{\sqrt{n}} = \frac{2}{10} = 0.2$$

Step 3 — express the probability with c.d.f.s. The range $[19.8, 20.2]$ is exactly $\mu\pm0.2 = \mu\pm 1\,\mathrm{SE}$:

$$\mathbb{P}(19.8\le\bar X\le 20.2) = F_{\bar X}(20.2) - F_{\bar X}(19.8)$$

Since this is one standard error either side of the mean, the empirical rule gives it directly:

≈ 0.683

In lecture your professor evaluates the two c.d.f. terms numerically. On the exam the numbers will land on 1, 2, or 3 standard errors — or you will be asked to stop at the c.d.f. expression.

Step 1 — $\sigma$ is unknown, so substitute $S_X$.

$$\mathrm{SE}(\bar X) \approx \frac{S_X}{\sqrt{n}} = \frac{4}{\sqrt{64}} = \frac48 = 0.5$$

Step 2 — the CLT sampling distribution. $n=64\ge30$:

$$\bar{X}\;\dot\sim\;\mathcal{N}(10,\; 0.5^2)$$

Step 3 — how many standard errors below the mean is 9?

$$\frac{9-10}{0.5} = -2 \qquad\text{so } 9 = \mu - 2\,\mathrm{SE}$$

Step 4 — use symmetry. The central $\pm2$ region holds 0.954, so the two tails together hold $1-0.954 = 0.046$, and the lower tail alone holds half of that:

$$\mathbb{P}(\bar X\le 9) \approx \frac{0.046}{2}$$

≈ 0.023

The worksheet asked for this as code. The setup above — identify SE, count standard errors, halve the leftover — is the part that would be graded on a no-code exam.

(a) You are not told the distribution of an individual resistor, so this is a CLT problem. Check: $n=36\ge30$, i.i.d. (randomly selected), inferring the mean. Then

$$\mathbb{E}(\bar X) = \mu = 100 \qquad \mathbb{V}(\bar X) = \frac{\sigma^2}{n} = \frac{100}{36} = 2.78$$

X̄ ˙~ Normal(μ = 100, σ²/n = 100/36 ≈ 2.78)

Give the variance, not the standard deviation, when the distribution is written $\mathcal{N}(\mu,\sigma^2)$. The standard error is $\sqrt{2.78} = 10/6 \approx 1.667$.

(b) Express it in standard errors:

$$\frac{101-100}{1.667} = 0.6 \qquad\text{so the range is } \mu \pm 0.6\,\mathrm{SE}$$

$$\mathbb{P}(99\le\bar X\le101) = F_{\bar X}(101) - F_{\bar X}(99)$$

≈ 0.451

0.6 is not one of the memorized values, which is exactly why the homework told you to use two lines of Python. On a no-code exam, the setup above is the answer — identify the distribution, compute the standard error, and write the probability as a difference of c.d.f.s. Getting to that line is what earns the marks.

(a) $\mathbb{P}(A\cup B) = 0.60+0.50-0.35 = 0.75$

(b) "Exactly one" is $A\cup B$ minus the overlap: $0.75-0.35 = 0.40$. Or build the table: $\mathbb{P}(A\cap B^c) = 0.60-0.35 = 0.25$ and $\mathbb{P}(A^c\cap B)=0.50-0.35=0.15$, summing to 0.40.

(c) $\mathbb{P}(A^c\cap B^c) = 1-\mathbb{P}(A\cup B) = 0.25$

(d) Test it: $\mathbb{P}(A)\mathbb{P}(B) = (0.6)(0.5) = 0.30$, but $\mathbb{P}(A\cap B)=0.35$.

0.75 · 0.40 · 0.25 · not independent

Let $D$ = defective, $F$ = flagged. Given: $\mathbb{P}(D)=0.02$, $\mathbb{P}(F\mid D)=0.95$, $\mathbb{P}(F\mid D^c)=0.04$. You want $\mathbb{P}(D\mid F)$.

Denominator by L.o.T.P.:

$$\mathbb{P}(F) = (0.95)(0.02) + (0.04)(0.98) = 0.019 + 0.0392 = 0.0582$$

Bayes:

$$\mathbb{P}(D\mid F) = \frac{0.019}{0.0582}$$

≈ 0.326

A flagged chip is still probably fine. The sensor is good, but defects are rare, so the 4% false-alarm rate applied to 98% of chips produces more false flags than true ones. Recognizing this pattern is what Bayes questions are testing.

(a) $$\int_0^2 c(1+x)\,dx = c\left[x+\tfrac{x^2}{2}\right]_0^2 = c(2+2) = 4c = 1$$

c = 1/4

(b) $$\mathbb{E}(X) = \int_0^2 x\cdot\tfrac14(1+x)\,dx = \tfrac14\int_0^2 (x+x^2)\,dx = \tfrac14\left[\tfrac{x^2}{2}+\tfrac{x^3}{3}\right]_0^2 = \tfrac14\left(2+\tfrac83\right) = \tfrac14\cdot\tfrac{14}{3}$$

E(X) = 7/6 ≈ 1.167

Above the midpoint of 1, which is right — the density increases with $x$, so mass piles toward the upper end.

(c) Since $\{X>1\}\subset\{X>0.5\}$, the conditional is a plain ratio.

$$\mathbb{P}(X>1) = \tfrac14\left[x+\tfrac{x^2}{2}\right]_1^2 = \tfrac14\left(4 - 1.5\right) = \tfrac58$$

$$\mathbb{P}(X>0.5) = \tfrac14\left[x+\tfrac{x^2}{2}\right]_{0.5}^{2} = \tfrac14\left(4 - 0.625\right) = \tfrac{27}{32}$$

$$\frac{5/8}{27/32} = \frac58\cdot\frac{32}{27}$$

20/27 ≈ 0.741

Do $X$ first, then transform — do not rebuild a p.m.f. for $Y$.

$$\mathbb{E}(X) = 1(0.2)+2(0.5)+4(0.3) = 0.2+1.0+1.2 = 2.4$$

$$\mathbb{E}(X^2) = 1(0.2)+4(0.5)+16(0.3) = 0.2+2.0+4.8 = 7.0$$

$$\mathbb{V}(X) = 7.0-(2.4)^2 = 7.0-5.76 = 1.24$$

Now apply the rules with $a=-2$, $b=5$:

$$\mathbb{E}(Y) = -2(2.4)+5 = 0.2 \qquad \mathbb{V}(Y) = (-2)^2(1.24) = 4.96 \qquad \sigma_Y = |{-2}|\sqrt{1.24}$$

E(Y) = 0.2 · V(Y) = 4.96 · σ_Y ≈ 2.227

The negative $a$ flips the distribution but the variance is unaffected by the sign, because $a$ enters squared.

The number of trials is fixed at 12, so $X\sim\mathrm{Binom}(12, 0.08)$.

(a) $$p_X(2) = \binom{12}{2}(0.08)^2(0.92)^{10} = 66(0.0064)(0.4344)$$

≈ 0.183

$\binom{12}{2} = \frac{12!}{2!\,10!} = \frac{12\cdot11}{2} = 66$.

(b) $$\mathbb{P}(X\ge1) = 1-\mathbb{P}(X=0) = 1-(0.92)^{12} = 1-0.3677$$

≈ 0.632

(c) $$\mathbb{E}(X) = kp = 12(0.08) = 0.96 \qquad \mathbb{V}(X) = 12(0.08)(0.92) = 0.8832$$

E(X) = 0.96 · σ ≈ 0.940

Counts in a window with no upper bound — Poisson. Convert the rate first.

(a) For a 5-metre window, $\lambda = 0.3\times 5 = 1.5$:

$$\mathbb{P}(X=0) = e^{-1.5}$$

≈ 0.223

(b) $$\mathbb{P}(X>1) = 1 - p_X(0)-p_X(1) = 1 - e^{-1.5} - 1.5e^{-1.5} = 1 - e^{-1.5}(2.5) = 1-0.558$$

≈ 0.442

(c) For a span of length $L$, $\lambda = 0.3L$. Set up and solve:

$$\mathbb{P}(X\ge1) = 1-e^{-0.3L} = 0.9 \;\Longrightarrow\; e^{-0.3L} = 0.1 \;\Longrightarrow\; 0.3L = \ln 10$$

L = ln(10)/0.3 ≈ 7.68 metres

First: $\sigma^2 = 4$, so $\sigma = 2$. That step is worth a mark on its own.

(a) $46 = \mu-2\sigma$ and $54 = \mu+2\sigma$, so this is the central $\pm2\sigma$ band.

≈ 0.954

(b) Leftover $= 1-0.954 = 0.046$, split evenly between the two tails.

≈ 0.023

(c) $52 = \mu+\sigma$, so $\mathbb{P}(X>52) = (1-0.683)/2 = 0.1585$. Since $\{X>54\}\subset\{X>52\}$:

$$\frac{0.023}{0.1585}$$

≈ 0.145

(d) $\mu\pm1.960\sigma = 50\pm 1.960(2) = 50\pm3.92$

[46.08, 53.92] mm

"Mean 500 hours" means $\theta = 500$. Use $F_X(x) = 1-e^{-x/\theta}$.

(a) $$\mathbb{P}(X\le 200) = 1-e^{-200/500} = 1-e^{-0.4}$$

≈ 0.330

(b) $$\mathbb{P}(X>1000) = e^{-1000/500} = e^{-2}$$

≈ 0.135

(c) For an exponential the standard deviation equals the mean.

σ = 500 hours

(d) Now solve for the parameter:

$$e^{-200/\theta} = 0.9 \;\Longrightarrow\; -\frac{200}{\theta} = \ln(0.9) \;\Longrightarrow\; \theta = \frac{200}{-\ln 0.9} = \frac{200}{0.1054}$$

θ ≈ 1898 hours

(a) $$\mathcal{L}(\theta\mid\mathbf{x}) = \prod_{i=1}^{n}\theta^2 x_i e^{-\theta x_i} = \theta^{2n}\left(\prod_{i=1}^n x_i\right)e^{-\theta\sum_{i=1}^n x_i}$$

Three groups: the $\theta^2$ terms multiply to $\theta^{2n}$; the $x_i$ collect into a product; the exponents add.

(b) $$\ell = 2n\ln\theta + \sum_{i=1}^{n}\ln x_i - \theta\sum_{i=1}^{n}x_i$$

(c) Differentiate. The middle term has no $\theta$, so it disappears:

$$\frac{d\ell}{d\theta} = \frac{2n}{\theta} - \sum_{i=1}^{n}x_i = 0 \;\Longrightarrow\; \theta = \frac{2n}{\sum x_i}$$

θ̂_MLE = 2n / ∑x_i = 2/X̄

(d) Invariance: push $\hat\theta_{MLE}$ through $g(\theta)=2/\theta$.

$$\widehat{\text{mean}} = \frac{2}{\hat\theta_{MLE}} = \frac{2}{2/\bar X}$$

= X̄

A satisfying result: the MLE of the population mean turns out to be the sample mean. The "spot the term with no $\theta$ and drop it" move in (c) is the one that makes these problems fast.

Where each entry comes from. $\hat\theta_2$ is a sample mean of only two data, so its variance is $\sigma^2/2$ — throwing away half your data doubles the variance. For $\hat\theta_3$, use $\mathbb{E}(aX)=a\mathbb{E}(X)$ and $\mathbb{V}(aX)=a^2\mathbb{V}(X)$ with $a=0.9$, giving $0.81\times\sigma^2/4 = 0.2025\sigma^2$.

Which to choose. $\hat\theta_1$ beats $\hat\theta_2$ always: same bias, half the variance. Comparing $\hat\theta_1$ and $\hat\theta_3$:

$$0.01\mu^2 + 0.2025\sigma^2 < 0.25\sigma^2 \;\iff\; 0.01\mu^2 < 0.0475\sigma^2 \;\iff\; |\mu| < 2.18\,\sigma$$

θ̂₃ wins when |μ| < 2.18σ; otherwise θ̂₁

This is the whole point of MSE: a deliberately biased estimator can beat an unbiased one when it buys enough precision. In practice you would not use $\hat\theta_3$ because the condition depends on the $\mu$ you do not know — but the arithmetic shows why "unbiased" is not automatically "best."

(a) For a Poisson, the sufficient statistic for $\lambda$ is the total:

$$X_+ = \sum_{i=1}^{8}x_i = 3+5+2+4+6+3+4+5 = 32$$

(b) From the master table, $X_+\sim\mathrm{Pois}(n\lambda)$:

X₊ = 32,   X₊ ~ Pois(8λ)

(c) Run the five steps.

$$\mathcal{L}(\lambda\mid\mathbf{x}) = \prod_{i=1}^{n}\frac{\lambda^{x_i}e^{-\lambda}}{x_i!} = \frac{\lambda^{\sum x_i}e^{-n\lambda}}{\prod x_i!}$$

$$\ell = \left(\sum x_i\right)\ln\lambda - n\lambda - \ln\left(\prod x_i!\right)$$

The last term carries no $\lambda$, so it dies on differentiation:

$$\frac{d\ell}{d\lambda} = \frac{\sum x_i}{\lambda} - n = 0 \;\Longrightarrow\; \hat\lambda_{MLE} = \frac1n\sum x_i = \bar X$$

$$\hat\lambda = \frac{32}{8}$$

λ̂_MLE = 4 calls per hour

(a) The distribution of a single task time is unknown, so the master table does not apply. But the data are i.i.d., $n = 64 \ge 30$, and you are inferring the mean — so the CLT applies.

$$\mathrm{SE}(\bar X) = \frac{\sigma}{\sqrt n} = \frac{12}{8} = 1.5$$

X̄ ˙~ Normal(45, 1.5² = 2.25)

(b) $48 = 45 + 2(1.5) = \mu + 2\,\mathrm{SE}$, so this is the upper $2\sigma$ tail of the sampling distribution:

$$\mathbb{P}(\bar X > 48) = \frac{1-0.954}{2}$$

≈ 0.023

(c) $$\frac{12}{\sqrt n} \le 1 \;\Longrightarrow\; \sqrt n \ge 12 \;\Longrightarrow\; n \ge 144$$

n = 144

Going from $\mathrm{SE}=1.5$ to $\mathrm{SE}=1$ costs you more than twice the data. That is the $\sqrt n$ at work.

Step 1 — name the events. $C$ = takes calculus, $S$ = takes statistics.

$$\mathbb{P}(C) = 0.45 \qquad \mathbb{P}(S) = 0.30 \qquad \mathbb{P}(C\cap S) = 0.20$$

Step 2 — write in what you know. "Both" = 0.20 goes top-left. $\mathbb{P}(C)$ = 0.45 goes at the bottom of the $C$ column. $\mathbb{P}(S)$ = 0.30 goes at the end of the $S$ row.

Step 3 — fill by subtraction, one cell at a time:

• Only $S$: the $S$ row totals 0.30, and 0.20 is already used. $\;0.30 - 0.20 = 0.10$
• Only $C$: the $C$ column totals 0.45, and 0.20 is already used. $\;0.45 - 0.20 = 0.25$
• $C^c$ column total: $\;1 - 0.45 = 0.55$
• Neither: the $C^c$ column totals 0.55, and 0.10 is already used. $\;0.55 - 0.10 = 0.45$

Check: $0.20 + 0.10 + 0.25 + 0.45 = 1.00$ ✓

Step 4 — read off each answer.

(a) At least one = everything except neither: $\;1 - 0.45$

0.55

(b) Exactly one = only $C$ + only $S$: $\;0.25 + 0.10$

0.35

(c) Neither = bottom-right cell.

0.45

(d) "Among students who take calculus" means shrink the world to the $C$ column. That column totals 0.45. The part of it that also takes statistics is the top-left cell, 0.20.

$$\mathbb{P}(S\mid C) = \frac{\mathbb{P}(C\cap S)}{\mathbb{P}(C)} = \frac{0.20}{0.45}$$

4/9 ≈ 0.444

(e) Multiply the two totals and compare with the top-left cell:

$$\mathbb{P}(C)\times\mathbb{P}(S) = 0.45\times 0.30 = 0.135 \qquad\text{but}\qquad \mathbb{P}(C\cap S) = 0.20$$

Not independent

Part (d) says the same thing a second way: 30% of all students take statistics, but 44% of calculus students do. Knowing someone takes calculus changes the odds, so the events are linked.

What makes this one harder: you are not given $\mathbb{P}(B)$. You get "neither" and "both" instead, and have to work out the rest.

Step 1 — name the events. $A$ = component A fails, $B$ = component B fails.

$$\mathbb{P}(A) = 0.15 \qquad \mathbb{P}(A^c\cap B^c) = 0.78 \qquad \mathbb{P}(A\cap B) = 0.05$$

Step 2 — write in what you know. "Both" = 0.05 goes top-left. "Neither" = 0.78 goes bottom-right. $\mathbb{P}(A)$ = 0.15 goes at the bottom of the $A$ column.

Step 3 — fill by subtraction. Start with the $A$ column, because you know its total:

• Only $A$: the $A$ column totals 0.15, and 0.05 is used. $\;0.15 - 0.05 = 0.10$
• $A^c$ column total: $\;1 - 0.15 = 0.85$
• Only $B$: the $A^c$ column totals 0.85, and 0.78 is used. $\;0.85 - 0.78 = 0.07$
• $B$ row total: add across the top row. $\;0.05 + 0.07 = 0.12$
• $B^c$ row total: add across the bottom row. $\;0.10 + 0.78 = 0.88$

Check: $0.05 + 0.07 + 0.10 + 0.78 = 1.00$ ✓

Step 4 — read off each answer.

(a) At least one = everything except neither: $\;1 - 0.78$

0.22

(b) The total of the $B$ row, which you just built.

0.12

(c) Exactly one = only $A$ + only $B$: $\;0.10 + 0.07$

0.17

(d) $$\mathbb{P}(A)\times\mathbb{P}(B) = 0.15\times 0.12 = 0.018 \qquad\text{but}\qquad \mathbb{P}(A\cap B) = 0.05$$

Not independent

Both components fail together almost three times as often as independence would predict. In a real machine that points to a shared cause, like a power surge that takes out both.

Step 1 — name the events. $D$ = has the condition (the cause). $T$ = tests positive (the evidence), so $T^c$ = tests negative.

Step 2 — translate each sentence.

Step 3 — the Bayes table, with "tests positive" as the evidence:

The 0.99 comes from $\mathbb{P}(D^c) = 1 - 0.01$.

(a) The total of the product column.

P(T) = 0.0585

(b) Your row's product divided by the total:

$$\mathbb{P}(D\mid T) = \frac{0.0090}{0.0585}$$

≈ 0.154

(c) The evidence is now "tests negative," so you need new likelihoods. Each one is a complement within the same condition:

$$\mathbb{P}(T^c\mid D) = 1 - \mathbb{P}(T\mid D) = 1 - 0.90 = 0.10$$

$$\mathbb{P}(T^c\mid D^c) = 1 - \mathbb{P}(T\mid D^c) = 1 - 0.05 = 0.95$$

$$\mathbb{P}(D\mid T^c) = \frac{0.0010}{0.9415}$$

≈ 0.001

Check: $\mathbb{P}(T^c) = 0.9415 = 1 - 0.0585$ ✓. A negative result is very reassuring (about 1 in 1,000), even though a positive result was not very alarming.

Step 1 — name the events. $M_1, M_2, M_3$ = which machine made the part (the causes). $D$ = the part is defective (the evidence).

Step 2 — translate. "Machine 1 makes 50%" gives $\mathbb{P}(M_1) = 0.50$. "A defect rate of 1%" for machine 1 means if it came from machine 1, it's defective 1% of the time: $\mathbb{P}(D\mid M_1) = 0.01$.

Step 3 — the Bayes table. Three causes means three rows:

The last column adds to 1 ✓ — a defective part came from some machine.

(a) The total of the product column. This is the law of total probability with three branches:

$$\mathbb{P}(D) = (0.50)(0.01) + (0.30)(0.03) + (0.20)(0.06) = 0.005 + 0.009 + 0.012$$

0.026

(b) Machine 3's product divided by the total:

$$\mathbb{P}(M_3\mid D) = \frac{0.012}{0.026}$$

≈ 0.462

(c) "Both" is an intersection, not a conditional. Read machine 3's product and don't divide:

$$\mathbb{P}(M_3\cap D) = \mathbb{P}(D\mid M_3)\,\mathbb{P}(M_3) = (0.06)(0.20)$$

0.012

(d) The largest value in the last column.

Machine 3 (46%)

Machine 3 makes the fewest parts, but its defect rate is so high that it produces almost half of all defects. Machine 1 makes half the parts but only 19% of the defects. The answer depends on both the prior and the likelihood, which is why you multiply them.