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[ASK] probability of that goalkeeper being able to fend off those penalty kicks 3 times

Monoxdifly

Well-known member
Aug 6, 2015
278
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?
 

tkhunny

Well-known member
MHB Math Helper
Jan 27, 2012
267
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?
Sometimes, if the distribution is small enough, it pays to calculate the entire Binomial Distribution and answer what may be a whole collection of questions.

1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1

0 Blocks: 1 (3/5)^0 (2/5)^5
1 Block: 5 (3/5)^1 (2/5)^4
2 Blocks:10 (3/5)^2 (2/5)^3
3 Blocks: 10 (3/5)^3 (2/5)^2
4 Blocks: 5 (3/5)^4 (2/5)^1
5 Blocks: 1 (3/5)^5 (2/5)^0

It's a pretty simple pattern to follow. Of course, if you wish, you can just calculate the desired value directly.
 

Monoxdifly

Well-known member
Aug 6, 2015
278
Ah, I think I get it. Thanks.