Question

mitchell plays college soccer. he makes a goal 50% of the time he shoots. mitchell is going to attempt two goals in a row in the next game. a = the event mitchell is successful on his first attempt. p(a) = 0.5. b = the event mitchell is successful on his second attempt. p(b) = 0.5. mitchell tends to shoot in streaks. the probability that he makes the second goal given that he made the first goal is 0.85. a. what is the probability that he makes both goals? b. what is the probability that mitchell makes either the first goal or the second goal?

Explanation:

Step1: Recall conditional - probability formula

The formula for conditional probability is $P(B|A)=\frac{P(A\cap B)}{P(A)}$, and we want to find $P(A\cap B)$. We know that $P(A) = 0.5$ and $P(B|A)=0.85$.

Step2: Solve for $P(A\cap B)$ for part a

Rearranging the conditional - probability formula gives $P(A\cap B)=P(B|A)\times P(A)$. Substituting the given values, we have $P(A\cap B)=0.85\times0.5 = 0.425$.

Step3: Recall the addition rule for probability for part b

The addition rule for probability is $P(A\cup B)=P(A)+P(B)-P(A\cap B)$. We know that $P(A) = 0.5$, $P(B)=0.5$, and $P(A\cap B)=0.425$.

Step4: Calculate $P(A\cup B)$

Substitute the values into the addition - rule formula: $P(A\cup B)=0.5 + 0.5-0.425=0.575$.

Answer:

a. $0.425$
b. $0.575$