If two dice are thrown simultaneously then find the probability that the sum of the numbers coming up on them is 11, given that the number 5 always occurs on the first dice.
1. 1/3
2. 1/36
3. 1/12
4. 1/6
1. 1/3
2. 1/36
3. 1/12
4. 1/6
Correct Answer – Option 4 : 1/6
Concept:
P(A \(\cap\) B) = P(A) x P(B | A) = P(B) x P(A | B) where P(A | B) represents the conditional probability of A given B and P (A | B) represents the conditional probability of B given A.
Calculation:
Let S be the sample space
∴ n(S) = 36
Let, A = the event that the sum of the numbers on the two dice is 11.
∴ A = {(5,6), (6,5)}
∴ n(A) = 2
Let, B = the event of the occurrence of 5 on the first dice.
B = {(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)}
∴ n(B) = 6
Now, P(B) = n(B) / n(S) = 6/36 = 1/6.
⇒ A ∩ B = {(5,6)}
∴ n(A∩B) = 1
Now, P(A ∩ B) = n(A ∩ B) / n(S) = 1/36.
⇒ P(A | B) = P(A ∩ B) / P(B) = 1/6
Hence, option 4 is correct.