# One ticket is selected at random from 50 tickets numbered 00, 01, 02, .... 49.

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One ticket is selected at random from 50 tickets numbered 00, 01, 02, .... 49.Then the probability that the sum of the digits selected is 8, given that the product of these digits is zero is equal to

(a) 1/14

(b) 1/7

(c) 5/14

(d) 1/50

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Answer: (a)$\frac{1}{14}$

Let S = {00, 01, 02, ... , 48, 49}

n(S) = 50

Let A be the event that the sum of the digits on the selected ticket is 8, then

A = {08, 17, 26, 35, 44} ⇒ n(A) = 5

Let B be the event that the product of the digits is zero.

Then, B = {00, 01, 02, ... , 08, 09, 10, 20, 30, 40}

⇒ n(B) = 14

∴  A ∩ B = {08} ⇒ n(A ∩ B) = 1

∴  Required probability = $P\big(\frac{A}{B}\big)$

$\frac{P(A\,\cap\,B)}{P(B)} =\frac{1/150}{14/50}$ = $\frac{1}{14}$