Probability of goalkeeper saving penalties

[RBS_5]

I am having a problem with one of my Probability homework questions.

A goalkeepr expects to save one penalty out of every three. Calculate the probability that he :

1- Saves one penalty out of the next three,

2- Fails to save any of the next three penalties,

3- Saves two out of the next three penalties.

Help.

 

[ShawnD]

A goalkeepr expects to save one penalty out of every three. Calculate the probability that he :

1- Saves one penalty out of the next three,

2- Fails to save any of the next three penalties,

3- Saves two out of the next three penalties.

IIRC, the method was
probability = (probability of yes)^(# of yes) * (probability of no)^(# of no)

For 1 it would be

$$(\frac{1}{3})^1 * (\frac{2}{3})^2$$

= 0.148


I could be wrong though.

 

[cookiemonster]

Shawn's close, but you forgot to consider the order. That is indeed the probability that he will make one save and two misses, but he could do that in 3 different ways, i.e. make the save on the first or the second or the third.

What you're looking for is

$$\binom{n}{m}p^m(1-p)^{n-m}$$

if p is the probability of success, n is the number of tries, and m is the number of successes.

cookiemonster

 

[RBS_5]

Thanks , that was a big help.