Q. 85.0( 1 Vote )

# Two dice are thrown together. The probability that neither they show equal digits nor the sum of their digits is 9 will beA. B. C. D. In a single throw of 2 die, we have total 36(6 × 6) outcomes possible.

Say, n(S) = 36 where S represents Sample space

Let A denotes the event of getting a doublet(equal number)

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

P(A) = And B denotes the event of getting a total of 9

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

P(B) = We need to find probability of the event of getting neither a doublet nor a total of 9.

P(A’ B’) = ?

As, P(A’ B’) = P(A B)’ {using De Morgan’s theorem}

P(A’ B’) = 1 – P(A B)

Note: By definition of P(E or F) under axiomatic approach(also called addition theorem) we know that:

P(E F) = P(E) + P(F) – P(E F)

P(A B) = {As P(A B) = 0 since nothing is common in set A and B n(A B) = 0 }

Hence,

P(A’ B’) = 1 – (5/18) = 13/18

Hence,

P(required event) = 13/18

As our answer matches only with option (b)

Option (b) is the only correct choice.

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation view all courses 