Question

Two number cubes are rolled for two separate events:

Answer 1

Answer:  The answer is $\dfrac{11}{28}.$

Step-by-step explanation: Given that two number cubes are rolled for two separate events A and B, where

A = the event that the sum of numbers on both cubes is less than 10

  = {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5) ,(5,1), (5,2), (5,3), (5,4),  (6,1), (6,2), (6,3)}

and

B =the event that the sum of numbers on both cubes is a multiple of 3

= {(1,2), (1,5),  (2,1), (2,4),  (3,3), (3,6), (4,2), (4,5), (5,1), (5,4), (6,3), (6,6) }

Let 'S' be the sample space for the experiment so that n(S) = 36.

Therefore, A and B = The event that the sum of numbers on both the cubes    is less than 10 and multiple of 3

   = {(1,2),  (1,5), (2,1), (2,4),  (3,3), (3,6), (4,2), (4,5) ,(5,1), (5,4), (6,3)}.

So, n(S) = 36,  n(A) = 30,  n(B) = 12  and n(A and B) = 11.

Hence, the conditional probability for the event B given that event A has already occured is given by

$P(B/A)=\dfrac{P(B\cap A)}{P(A)},$

where,

$P(B\cap A)=\dfrac{n(B\cap A}{n(S)}=\dfrac{11}{36}.$

$P(A)=\dfrac{n(A)}{n(S}=\dfrac{28}{36}.$

Therefore,

$P(B/A)=\dfrac{\frac{11}{36}}{\frac{28}{36}}=\dfrac{11}{28}.$

Thus, the required conditional probability is $\dfrac{11}{28}.$