MHB Probability of Keeper Fending Off Penalty Kicks 3/5 Times

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A professional goalkeeper has a probability of 3/5 to fend off a penalty kick. In a scenario where five penalty kicks are taken, the probability of the goalkeeper successfully saving three of them can be calculated using the Binomial Distribution. The discussion highlights the importance of understanding the distribution to solve the problem effectively. Participants suggest calculating the entire distribution or directly finding the desired value. The conversation emphasizes the straightforward nature of the calculations involved.
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A professional goalkeeper can fend off a penalty kick with the probability of $$\frac{3}{5}$$. 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. $$\frac{180}{625}$$
B. $$\frac{612}{625}$$
C. $$\frac{216}{625}$$
D. $$\frac{228}{625}$$
E. $$\frac{230}{625}$$

Can someone give me a hint?
 
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Monoxdifly said:
A professional goalkeeper can fend off a penalty kick with the probability of $$\frac{3}{5}$$. 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. $$\frac{180}{625}$$
B. $$\frac{612}{625}$$
C. $$\frac{216}{625}$$
D. $$\frac{228}{625}$$
E. $$\frac{230}{625}$$

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.
 
Ah, I think I get it. Thanks.
 
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