## Solution :

S = { 00, 01, 02, ……, 49 }

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

A = { 08, 17, 26, 35, 44 }

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

B = { 00, 01, 02, 03, …. , 09, 10, 20, 30, 40 }

\(\therefore\) \(A \cap B\) = { 8 }

\(\therefore\) Required probability = \(P({A\over B})\) = \(P(A \cap B)\over P(B)\)

= \(1/50\over 14/50\) = \(1\over 14\)

### Similar Questions

The probability of India winning a test match against the west indies is 1/2 assuming independence from match to match. The probability that in a match series India’s second win occurs at the third test is

A fair die is tossed eight times. The probability that a third six is observed on the eight throw, is

If A and B are two mutually exclusive events, then

It is given that the events A and B are such that P(A) = 1/4, P(A/B) = 1/2 and P(B/A) = 2/3. Then, P(B) is equal to

A die is thrown. Let A be the event that the number obtained is greater than 3. Let B be the event that the number obtained is less than 5. Then, \(P(A \cup B)\) is