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

#### Monoxdifly

##### Well-known member
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
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
Ah, I think I get it. Thanks.