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
[tex]P(A \ \text{ and } B) = P(A) \cdot P(B)[/tex]
=> [tex]40\% \times 5\%[/tex]
=> [tex]\frac{40}{100} \times \frac{5}{100}[/tex]
=>[tex] 0.4 \times 0.05[/tex]
=>[tex]0.02[/tex]
=>[tex]2\%[/tex]
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
=> [tex]1 - 2\%[/tex]
=> [tex]1 - 0.02[/tex]
=> [tex]0.98[/tex]
=>[tex]98\%[/tex]
