Question

1. (4 points) let a and b be events. suppose that the probability that a occurs is .5, the probability that b does not occur is .6, and the probability that either a or b occurs is .7. (a) find the conditional probability that a occurs given that b does not occur. (b) find the conditional probability that a does not occur given that b occurs. 5) (4 points) suppose there are 80 participants in a contest. find the number of ways in which three different prizes - the first, second, and third - can be awarded if no contestant can win more than one prize.

Explanation:

Step1: Find relevant probabilities

We know $P(A) = 0.5$, $P(\overline{B})=0.6$, so $P(B)=1 - P(\overline{B}) = 1- 0.6=0.4$, and $P(A\cup B)=0.7$. Using the formula $P(A\cup B)=P(A)+P(B)-P(A\cap B)$, we can find $P(A\cap B)$. So $0.7 = 0.5+0.4 - P(A\cap B)$, then $P(A\cap B)=0.5 + 0.4-0.7=0.2$. Also, $P(A\cap\overline{B})=P(A)-P(A\cap B)=0.5 - 0.2 = 0.3$.

Step2: Solve part (a)

The formula for conditional - probability is $P(A|\overline{B})=\frac{P(A\cap\overline{B})}{P(\overline{B})}$. Substituting the values we found, $P(A|\overline{B})=\frac{0.3}{0.6}=0.5$.

Step3: Solve part (b)

First, we find $P(\overline{A}\cap B)=P(B)-P(A\cap B)=0.4 - 0.2=0.2$. Then, using the conditional - probability formula $P(\overline{A}|B)=\frac{P(\overline{A}\cap B)}{P(B)}$. Substituting the values, $P(\overline{A}|B)=\frac{0.2}{0.4}=0.5$.

Step4: Solve problem 5

The number of ways to award 3 different prizes to 80 participants with no contestant winning more than one prize is a permutation problem. The formula for permutations is $P(n,r)=\frac{n!}{(n - r)!}$, where $n = 80$ and $r = 3$. So $P(80,3)=\frac{80!}{(80 - 3)!}=\frac{80!}{77!}=80\times79\times78=492960$.

Answer:

(a) $0.5$
(b) $0.5$

1. $492960$