Assume that there are a sequence of consecutive integers 1, 2, 3, 4, 5, ... 15....

Assume that there are a sequence of consecutive integers 1, 2, 3, 4, 5, ... 15. Tom and Jim respectively select a number from the sequence randomly (no repetition). Given that Tom’s number is divisible by 5, what’s the probability that Tom’s number is greater than Jim’s number ?

Expert Solution

A = Event that Tom's number is greater than Jim's number.

B = Event that Tom's number is divisible by 5.

So, we have to find P ( A|B) = P ( A B) / P ( B)

A B = Event that Tom's number is divisible by 5 and it is greater than Jim's number

= (Tom's number is 5 , Jim's number is any number from 1 to 4) ( Tom's number is 10 , Jim's number is any number from 1 to 9) ( Tom's number is 15 , Jim's number is any number from 1 to 14)

Total number of cases in A B is = 4+9+14 = 27

Tom select any number in 15 ways and for every 15 numbers of Tom, Jim can select numbers in 14 was. Total cases = 15*14 = 210

So, P(A B )= 27 / 210

P(B) = P( Tom select 5 or 10 or 15 from list of 15 numbers ) = 3 /15 =1/5

So,

P ( A|B) = (27/210) / ( 1/5 ) = 9/14 or 0.643

Given that Tom’s number is divisible by 5, the probability that Tom’s number is greater than Jim’s number is 9/14 or 0.643

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orchestra answered 2 years ago

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