Find the probability that the sum of the numbers showing on two dice i

1.	Find the probability that the sum of the numbers showing on two dice is 8, given that least one die does not show five.
Answer» We know, when a pair of dice is thrown, total possible outcomes are = 36 A=No of outcomes for getting sum of 8 are {(2,6),(3,5),(4,4),(5,3),(6,2)} = 5 Therefore, P(A) = \(\cfrac5{36}\) B = No of outcomes for at least one die does not show 5 = {(1,1),(1,2),(1,3),(1,4),(1,6),(2,1),(2,2),(2,3),(2,4),(2,6),(3,1),(3,2),(3,3),(3,4),(3,6),(4,1),(4,2), (4,3),(4,4),(4,6),(6,1),(6,2),(6,3),(6,4),(6,6) } = 25 Hence P(B) = \(\cfrac{25}{36}\) (A B) = outcomes of getting sum as 8 with at least one die not showing 5 = {(2,6),(4,4),(6,2)= 3 Hence P(A B) = \(\cfrac3{36}\) Therefore, \(P(\cfrac{A}{B})=\cfrac{P(A\cap B)}{P(B)}\) = \(\cfrac{\frac3{36}}{\frac{25}{36}}=\cfrac3{25}\)

Find the probability that the sum of the numbers showing on two dice is 8, given that least one die does not show five.

Answer»

We know, when a pair of dice is thrown, total possible outcomes are = 36

A=No of outcomes for getting sum of 8 are

{(2,6),(3,5),(4,4),(5,3),(6,2)} = 5

Therefore, P(A) = \(\cfrac5{36}\)

B = No of outcomes for at least one die does not show 5

= {(1,1),(1,2),(1,3),(1,4),(1,6),(2,1),(2,2),(2,3),(2,4),(2,6),(3,1),(3,2),(3,3),(3,4),(3,6),(4,1),(4,2), (4,3),(4,4),(4,6),(6,1),(6,2),(6,3),(6,4),(6,6) } = 25

Hence P(B) = \(\cfrac{25}{36}\)

(A B) = outcomes of getting sum as 8 with at least one die not showing 5

= {(2,6),(4,4),(6,2)= 3

Hence P(A B) = \(\cfrac3{36}\)

Therefore, \(P(\cfrac{A}{B})=\cfrac{P(A\cap B)}{P(B)}\)

= \(\cfrac{\frac3{36}}{\frac{25}{36}}=\cfrac3{25}\)

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