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

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

Leave a Comment

Your email address will not be published.