# 10th Grade IGCSE Mathematics HELP!

1. May 13, 2004

### 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.

2. May 13, 2004

### ShawnD

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.

3. May 13, 2004

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.