5. In a certain school there are only three sports offered: football, basketball

Question

5. In a certain school there are only three sports offered: football, basketball, and soccer. • For each of the three sports, the probability is 0.1 that a student only plays that sport. • For any two of the sports, the probability is 0.12 that a student plays these two sports but not the other • The probability that a student plays all three sports, given that he plays football and basketball is

1/ 3 What is the probability that a student takes none of the sports, given that he does not take football?

Explanation / Answer

Here we are given that :

P( football only ) = P( basketball only ) = P( soccer only ) = 0.1

P( football and basketball only ) = P( basketball and soccer only ) = P( soccer and football only ) = 0.12

P( all 3 sports | football and basketball ) = 1/3

Using Bayes theorem we get:

P( all 3 sports | football and basketball ) = P( all 3 games ) / P( basketball and football ) = 1/3

Therefore we get:

P( all 3 games ) / [ P ( all 3 games ) + P( basketball and football only ) ] = 1/3

Now putting P( basketball and football only ) = 0.12, we get:

P( all 3 games ) / [ 0.12 + P( all 3 games ) ] = 1/3

Cross multiplying we get:

0.12 + P( all 3 games ) = 3P(all 3 games )

Therefore, we get: P( all 3 games ) = 0.12 / [ 3-1 ] = 0.06

Therefore we get P( all 3 games ) = 0.06

Now the probability that the student takes at least one game is computed as:

= P( football only ) + P( basketball only ) + P( soccer only ) + P( football and basketball only ) + P( basketball and soccer only ) + P( football and soccer only ) + P( all 3 games )

= 0.1 + 0.1 + 0.1 + 0.12 + 0.12 + 0.12 + 0.06

= 0.72

Probability that the student takes none of the sports

= 1 - Probability that it takes at least one of the sports

= 1 - 0.72

= 0.28

Probability that a student does not play football

= 1 - P( football only ) - P( football and basketball only ) - P( football and soccer only ) - P( all 3 games )

= 1 - 0.1 - 0.12 - 0.12 - 0.06

= 0.6

Now given that he does not play foorball, probability that a student takes none of the sports is computed as:

= Probability that students takes none of the sports / Probability that he does not take football

= 0.28 / 0.6

= 0.4667

Therefore 0.4667 is the required probability here.

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