10th Grade IGCSE Mathematics HELP

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The discussion revolves around calculating the probabilities related to a goalkeeper's performance in saving penalties, specifically for three scenarios: saving one penalty, failing to save any, and saving two out of three. The initial method proposed for calculating the probability of saving one penalty was incorrect as it did not account for the different possible orders of saves and misses. The correct formula involves using combinations to account for the various ways the saves can occur, expressed as binomial probability: \binom{n}{m}p^m(1-p)^{n-m}. Participants clarified the calculations and provided guidance on the correct approach to solving the problems. This exchange highlights the importance of considering order in probability calculations.
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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. :smile:
 
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RBS_5 said:
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.
 
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
 
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Thanks , that was a big help.
 
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