Two number cubes are rolled for two separate events: event a is the event that the sum of numbers on both cubes is less than 10. event b is the event that the sum of numbers on both cubes is a multiple of 3. complete the conditional-probability formula for event b given that event a occurs first by writing a and b in the blanks: p( a0 | a1) = p( a2 ∩ a3) p( a4)

 The answer is

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

where,

Therefore,

Thus, the required conditional probability is

the answer is c.

for continuous functions, they continue in the same pattern and you have to find that pattern kinda like slope