1.

In a single throw of two dice, find the probability that neither a doublet nor a total of 9 will appear.

Answer»

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. 

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

∴ P(A) = \(\frac{n(A)}{n(S)}\) = \(\frac{6}{36}=\frac{1}{6}\) 

And B denotes the event of getting a total of 9 

∴ B = {(3,6), (6,3), (4,5), (5,4)}

P(B) = \(\frac{n(B)}{n(S)}\) = \(\frac{4}{36} = \frac{1}{9}\) 

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) = \(\frac{1}{6} +\frac{1}{9}+0\) = \(\frac{5}{18}\) 

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

Hence, 

P(A’∩B’) = 1 – \(\Big(\frac{5}{18}\Big)\)  = \(\frac{13}{18}\)


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