A dice is thrown two times and the sum of the scores appearing on the die is observed to be a multiple of 4. Then the conditional probability that the score 4 has appeared atleast once is :

Solution

Maths sir · 3 steps

Shrink the sample space to sums 4, 8 and 12, then count how many of those pairs show a 4.

1. New sample space. We are told the sum is a multiple of 4. With two dice the sum runs from 2 to 12, so it must be 4, 8 or 12. Only these outcomes now count.

   Sum	Ordered outcomes	Count	Pairs with a 4
   4	(1,3), (2,2), (3,1)	3	none
   8	(2,6), (3,5), (4,4), (5,3), (6,2)	5	(4,4)
   12	(6,6)	1	none

2. Count both events. The reduced sample space has 3 + 5 + 1 = 9 equally likely outcomes. Only (4,4) has a 4 in it, because a 4 on one die needs 0, 4 or 8 on the other, and only 4 is possible.

   P(at least one 4 | sum is a multiple of 4) = 1/9

3. So. The conditional probability is 1/9, which is option (b).

The idea to keep. In conditional probability, the given condition becomes your whole sample space. List those outcomes as ordered pairs, then count the favourable ones inside it.

Answer from the NTA final answer key. Solution by PrepWiser.