Conditional probability

$P(\text{basketball}) = \dfrac{12}{30} = \dfrac{2}{5} = 0.4$ .
$P(\text{swim} \mid \text{basketball}) = \dfrac{5}{12} \approx 0.417$ .
After removing a red marble: $5$ red and $4$ blue remain out of $9$ . $P(\text{red on 2nd draw}) = \dfrac{5}{9}$ .
$P(A \mid B) = \dfrac{P(A \text{ and } B)}{P(B)} = \dfrac{0.15}{0.5} = 0.3$ .
Yes, $A$ and $B$ are independent because $P(A \mid B) = 0.3 = P(A)$ . Knowing $B$ occurred does not change the probability of $A$ .
$P(\text{pet} \mid \text{male}) = \dfrac{20}{20 + 15} = \dfrac{20}{35} = \dfrac{4}{7} \approx 0.571$ .
$P(\text{male} \mid \text{pet}) = \dfrac{20}{20 + 25} = \dfrac{20}{45} = \dfrac{4}{9} \approx 0.444$ .
Yes, they are independent. The outcome of the coin does not affect the die, and vice versa. $P(\text{heads and 6}) = \dfrac{1}{2} \times \dfrac{1}{6} = \dfrac{1}{12} = P(\text{heads}) \times P(\text{6})$ .
After removing one heart, $12$ hearts remain out of $51$ cards. $P(\text{2nd heart} \mid \text{1st heart}) = \dfrac{12}{51} = \dfrac{4}{17}$ .
$P(A \mid B) = \dfrac{P(A \text{ and } B)}{P(B)}$ , where $P(B) > 0$ .

First draw: $P(G) = \dfrac{5}{8}$ , $P(Y) = \dfrac{3}{8}$ . If 1st is green: $P(G) = \dfrac{4}{7}$ , $P(Y) = \dfrac{3}{7}$ . If 1st is yellow: $P(G) = \dfrac{5}{7}$ , $P(Y) = \dfrac{2}{7}$ . $P(\text{both green}) = \dfrac{5}{8} \times \dfrac{4}{7} = \dfrac{20}{56} = \dfrac{5}{14}$ .
(a) $P(\text{German} \mid \text{French}) = \dfrac{80}{240} = \dfrac{1}{3} \approx 0.333$ . (b) $P(\text{French} \mid \text{German}) = \dfrac{80}{180} = \dfrac{4}{9} \approx 0.444$ .
(a) $P(\text{supports} \mid \text{under 30}) = \dfrac{45}{75} = 0.6$ . $P(\text{supports} \mid \text{30 and over}) = \dfrac{35}{75} \approx 0.467$ . (b) Yes, there is an association. The conditional probabilities differ: younger respondents are more likely to support the policy ( $60\%$ vs $47\%$ ). If there were no association, both groups would have the same support rate of $\dfrac{80}{150} \approx 0.533$ .
$P(A \text{ and } B) = P(A \mid B) \times P(B) = 0.75 \times 0.4 = 0.3$ . For independence: $P(A) \times P(B) = 0.6 \times 0.4 = 0.24$ . Since $0.3 \neq 0.24$ , $A$ and $B$ are not independent.
$P(\text{defective}) = 0.50 \times 0.02 + 0.30 \times 0.03 + 0.20 \times 0.05 = 0.010 + 0.009 + 0.010 = 0.029$ . The probability that a randomly selected item is defective is $2.9\%$ .

$P(\text{Z and defective}) = 0.20 \times 0.05 = 0.010$ . $P(\text{Z} \mid \text{defective}) = \dfrac{P(\text{Z and defective})}{P(\text{defective})} = \dfrac{0.010}{0.029} \approx 0.345$ . There is about a $34.5\%$ chance the defective item came from Machine Z.
Example: $P(\text{disease} \mid \text{positive test}) \neq P(\text{positive test} \mid \text{disease})$ . A test might detect $95\%$ of sick people ( $P(T^+ \mid D) = 0.95$ ), but if the disease is rare, $P(D \mid T^+)$ could be much lower (e.g. $0.16$ ). Confusing the two — called the “prosecutor’s fallacy” in legal contexts — leads to wildly wrong conclusions. In medicine, it means overestimating how likely a patient is to have the disease after a positive test.
$P(\text{1st blue}) = \dfrac{6}{10}$ . $P(\text{2nd blue} \mid \text{1st blue}) = \dfrac{5}{9}$ . $P(\text{3rd blue} \mid \text{first two blue}) = \dfrac{4}{8} = \dfrac{1}{2}$ . $P(\text{all three blue}) = \dfrac{6}{10} \times \dfrac{5}{9} \times \dfrac{1}{2} = \dfrac{30}{180} = \dfrac{1}{6}$ .
$P(A \text{ and } B) = P(A \mid B) \times P(B) = 0.5 \times 0.4 = 0.2$ . Also $P(A \text{ and } B) = P(B \mid A) \times P(A)$ , so $0.2 = 0.25 \times P(A)$ , giving $P(A) = \dfrac{0.2}{0.25} = 0.8$ .

Label the doors $1, 2, 3$ . Suppose the prize is behind door $1$ (by symmetry, the argument is the same for any door). You pick door $1$ : host opens door $2$ or $3$ ; switching loses. You pick door $2$ : host must open door $3$ ; switching wins. You pick door $3$ : host must open door $2$ ; switching wins. So switching wins $2$ out of $3$ times. Formally: let $W$ = prize behind your door. $P(W) = \dfrac{1}{3}$ . Given the host reveals a losing door, $P(\text{prize behind other door} \mid \text{host reveals}) = \dfrac{2}{3}$ .
$P(\text{uses drug}) = 0.005$ . $P(T^+ \mid \text{user}) = 0.99$ . $P(T^+ \mid \text{non-user}) = 0.02$ . $P(T^+) = 0.005 \times 0.99 + 0.995 \times 0.02 = 0.00495 + 0.0199 = 0.02485$ . $P(\text{user} \mid T^+) = \dfrac{0.00495}{0.02485} \approx 0.199$ . Only about $20\%$ of positive results are true positives. The test produces many false positives because the user base is so small. A confirmatory (more specific) test is essential.
$P(A \text{ and } \overline{B}) = P(A) - P(A \text{ and } B) = P(A) - P(A) \cdot P(B)$ (using independence) $= P(A)(1 - P(B)) = P(A) \cdot P(\overline{B})$ . Since $P(A \text{ and } \overline{B}) = P(A) \cdot P(\overline{B})$ , events $A$ and $\overline{B}$ are independent.
$P(\text{all 5 spades}) = \dfrac{13}{52} \times \dfrac{12}{51} \times \dfrac{11}{50} \times \dfrac{10}{49} \times \dfrac{9}{48} = \dfrac{13 \times 12 \times 11 \times 10 \times 9}{52 \times 51 \times 50 \times 49 \times 48} = \dfrac{154440}{311875200} = \dfrac{33}{66640} \approx 0.000495$ .

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