Probability and Decision Making

1. (a) The sample space has 52 outcomes — one for each card.

(b) There are 13 hearts:

$P(\text{heart}) = \frac{13}{52} = \frac{1}{4}$

(c) There are 3 face card ranks $\times$ 4 suits = 12 face cards:

$P(\text{face card}) = \frac{12}{52} = \frac{3}{13}$

(d) Cards that are both heart AND face card (Jack, Queen, King of hearts): 3 cards.

$P(\text{heart} \cup \text{face card}) = \frac{13}{52} + \frac{12}{52} - \frac{3}{52} = \frac{22}{52} = \frac{11}{26}$

2. (a) Check independence: $P(A) \cdot P(B) = 0.4 \times 0.3 = 0.12$.

Since $P(A \cap B) = 0.12 = P(A) \cdot P(B)$, events $A$ and $B$ are independent.

(b) Addition Rule:

$P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.4 + 0.3 - 0.12 = 0.58$

(c) Conditional probability:

$P(A \mid B) = \frac{P(A \cap B)}{P(B)} = \frac{0.12}{0.3} = 0.4$

Note: $P(A \mid B) = P(A)$ — as expected for independent events, knowing $B$ occurred doesn't change the probability of $A$.

3. (a) $P(\text{Grade A} \mid \text{studies daily}) = \dfrac{80}{100} = 0.80$

(b) $P(\text{Grade A} \mid \text{does not study daily}) = \dfrac{30}{100} = 0.30$

(c) Yes — the conditional probabilities are very different: $80\%$ of daily studiers earn Grade A versus only $30\%$ of non-daily studiers. This strong difference indicates an association between studying daily and earning Grade A. (This is correlation, not necessarily causation.)

4. Let $D$ = defective, $G$ = good, $T^+$ = positive test.

Given: $P(D) = 0.03$, $P(G) = 0.97$, $P(T^+ \mid D) = 0.90$, $P(T^+ \mid G) = 0.05$.

(a) $P(T^+ \cap D) = P(T^+ \mid D) \cdot P(D) = 0.90 \times 0.03 = 0.027$

(b) $P(T^+ \cap G) = P(T^+ \mid G) \cdot P(G) = 0.05 \times 0.97 = 0.0485$

(c) Total probability of a positive test:

$P(T^+) = P(T^+ \cap D) + P(T^+ \cap G) = 0.027 + 0.0485 = 0.0755$

(d) Conditional probability (Bayes' Theorem):

$P(D \mid T^+) = \frac{P(T^+ \cap D)}{P(T^+)} = \frac{0.027}{0.0755} \approx 0.358 \approx 36\%$

Even though a bulb tested positive, there is only about a 36% chance it is actually defective. This illustrates why false positives can dominate when the base rate of the condition is low.

5. (a) If the contestant stays with door $\#1$:

The probability the car was behind door $\#1$ when originally chosen is $\frac{1}{3}$. Staying wins only if the initial pick was correct:

$P(\text{win by staying}) = \frac{1}{3}$

(b) If the contestant switches to door $\#2$:

Switching wins if the car was originally behind door $\#2$ or $\#3$. Since the host always reveals a goat behind one of the other doors, the probability concentrates:

$P(\text{win by switching}) = \frac{2}{3}$

(c) Initially, the car is equally likely behind any door: $P(\text{car at \#1}) = P(\text{car at \#2}) = P(\text{car at \#3}) = \frac{1}{3}$.

The contestant chose door $\#1$. The host's action carries information: the host opens door $\#3$ (a goat), which is guaranteed regardless of whether the car is at $\#1$ or $\#2$.

• If car is at $\#1$ (prob $\frac{1}{3}$): staying wins, switching loses.
• If car is at $\#2$ (prob $\frac{1}{3}$): host must open $\#3$; switching wins.
• If car is at $\#3$ (prob $\frac{1}{3}$): host must open $\#2$; but host opened $\#3$, so this scenario is eliminated.

After door $\#3$ is opened, the $\frac{1}{3}$ probability that was on $\#3$ shifts entirely to door $\#2$. So $P(\text{car at \#2} \mid \text{host opens \#3}) = \frac{2}{3}$. Switching is twice as likely to win.

6. (a) Expected value:

$E = P(\text{success}) \times 500{,}000 + P(\text{failure}) \times (-200{,}000)$ $E = 0.6(500{,}000) + 0.4(-200{,}000)$ $E = 300{,}000 - 80{,}000 = 220{,}000$

The expected value of launching is 220,000 dollars.

(b) The expected value is positive, so launching is favorable based on expected value alone. However, other factors include: risk tolerance (a potential 200,000 dollar loss may be unacceptable), cash flow, competition, production costs, and whether the research sample was representative.

(c) New expected value with $P(\text{success}) = 0.75$:

$E = 0.75(500{,}000) + 0.25(-200{,}000) = 375{,}000 - 50{,}000 = 325{,}000$

The new expected value is $325{,}000 - 220{,}000 = 105{,}000$ dollars higher than before. Since the expected gain from the research (105,000 dollars) exceeds the cost (50,000 dollars), the research is worth purchasing.