Can someone help me please????

2 answers:

7 0

Let X be the random variable denoting the number of successful throws.

Here X~ Binomial Distribution with n = 5 and p = 0.80.

the probability of her missing 3 (or more) free throws out of 5

= P ( X ≤ 2)

= P (X= 0) + P(X= 1) + P(X= 2)

=0.00032 + 0.0064 + 0.0512
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= 0.05792

I hope my answer has come to your help. God bless and have a nice day ahead!
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zysi [14]4 years ago

4 0

Hello there.
<span>
Can someone help me please????

A basketball player makes 80% of her free throws. Recently during a very close game, she shot 5 free throws near the end of the game and missed 3 of them. The fans booed. What is the probability of her missing 3 (or more) free throws out of 5? Set up and conduct a simulation (using the random digits below) with 10 repetitions.
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0.05792

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