Answer :
Probability that Margie's first serve will be good at least four of the next six times she serves is 0.68
What is Binomial Distribution in Probability?
Assume that n independent replications of a random experiment with exactly two outcomes are performed. Success has a probability of [tex]p[/tex], while failure has a probability of [tex]q[/tex]. Assume that out of these n times, we will experience success [tex]x[/tex] times and failure [tex]n-x[/tex] times. There are [tex]^nC_x[/tex] different ways in which we can succeed. If, [tex]X[/tex] has a binomial distribution, then,
[tex]P(X=x)=p(x)[/tex]
[tex]=[/tex] [tex]^nC_xp^xq^{n-x}[/tex]
for [tex]x[/tex] = 0, 1,..., n. Here, [tex]q=1-p[/tex]. A binomial variate is any such random variable [tex]X[/tex]. A group of n separate Bernoullian trials is known as a binomial trial. Binomial distribution requirements:
- There are just two outcomes, success and failure, for each trial.
- The quantity "[tex]n[/tex]" of trials is finite.
- The trials run separately from one another.
For each trial, the odds of success [tex]p[/tex], or failure [tex]q[/tex], remain constant.
Given,
Probability of good serves = Success
⇒ [tex]p=\frac{20}{30}[/tex]
⇒[tex]p=\frac{2}{3}[/tex]
Failure = [tex]q[/tex]
⇒ [tex]q=1-p[/tex]
⇒ [tex]q=1-\frac{2}{3}[/tex]
⇒ [tex]q=\frac{1}{3}[/tex]
Number of trials [tex]= n[/tex]
[tex]n=6[/tex]
Probability of at least four good serve = [tex]P(4 < X < 6)[/tex]
According to the binomial distribution,
Therefore,
[tex]P(4 \leq X \leq 6)=P(X=4)+P(X=5)+P(X=6)[/tex]
[tex]=[/tex] [tex]^6C_4(\frac{2}{3}) ^4(\frac{1}{3})^2+ ^6C_5(\frac{2}{3}) ^5(\frac{1}{3})^1+^6C_6(\frac{2}{3}) ^6(\frac{1}{3})^0[/tex]
[tex]=[/tex] [tex]15\times\frac{16}{729} +6\times\frac{32}{729} +1\times\frac{64}{729}[/tex]
[tex]=[/tex] [tex]\frac{496}{729}[/tex]
[tex]= 0.68[/tex]
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