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A basketball player gets 2 free-throws shots when she is fouled by a player on the opposing team. She misses the first shot 40% of the time. When she misses. the first shot she misses the second shot by 5% of the time. What is the probability of missing tboth free throw shots?

Respuesta :

the answer to your question is 40/100 * 5/100
0.4 * 0.05
= 0.02
= 2%

Answer: There are 2% of the time that she missing the both free throw shots.

Step-by-step explanation:

Since we have given that

Probability of times she misses the first shot = 40%

Probability of times she misses the second shot = 5%

We are required to find the probability of missing the both free throw shots.

Since the above two shots are independent events,

So, we can apply the rule of independent events:

[tex]P(A\ and\ B)=P(A).P(B)\\\\\P(A\ and\ B)=\frac{40}{100}\times \frac{5}{100}\\\\P(A\ and\ B)=\frac{200}{10000}\\\\P(A\ and\ B)=0.02\times 100\\\\P(A\ and\ B)=2\%[/tex]

Hence, there are 2% of the time that she missing the both free throw shots.

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