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March 17th, 2018, 05:28 PM   #1
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[ASK] Professional Goalkeeper

A professional goalkeeper can fend off a penalty kick with the probability of $\displaystyle \frac{3}{5}$. In an event a kick is done 5 times. The probability of that goalkeeper being able to fend off those penalty kicks 3 times is ....
A. $\displaystyle \frac{180}{625}$
B. $\displaystyle \frac{612}{625}$
C. $\displaystyle \frac{216}{625}$
D. $\displaystyle \frac{228}{625}$
E. $\displaystyle \frac{230}{625}$

Can someone give me a hint?
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March 17th, 2018, 05:51 PM   #2
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Binomial distribution with parameters $n=5,~p=\dfrac 3 5$

$P[3] = \dbinom{5}{3}\left(\dfrac 3 5\right)^3 \left(\dfrac 2 5\right)^2 = \dfrac{216}{625}$, i.e. choice (c)
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March 18th, 2018, 01:37 AM   #3
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Okay, thank you.
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