Question

Please help with Part B

Answer 1

Answer:

The probability of missing both free-throw shots = 2%

The probability of making at least 1 free-throw shot = 98%

Step-by-step explanation:

Given :

Probability that she misses the first shot  = 40%

Probability she misses the second shot  = 5%

Solution:

Part A: The probability of missing both free-throw shots:

The two shots are independent events so, we can use rule of independent events

$P(A \ \text{ and } B) = P(A) \cdot P(B)$

=> $40\% \times 5\%$

=> $\frac{40}{100} \times \frac{5}{100}$

=>$0.4 \times 0.05$

=>$0.02$

=>$2\%$

Part B: The probability of making at least 1 free-throw shot

Probability of making at least 1 free throw show shot = 1 - probability of miss both the shots

=> $1 - 2\%$

=> $1 - 0.02$

=> $0.98$

=>$98\%$