If P ( E ) = 0 . 6 , P ( F ) = 0 . 3 and P ( E ∩ F ) = 0 . 2 , then find P E F .

Given P ( E ) = 0 . 6 , P ( F ) = 0 . 3 and P ( E ∩ F ) = 0 . 2 We know that, P E F = P ( E ∩ F ) P ( F ) ⇒ P E F = 0 . 2 0 . 3 = 2 3 Thus the value of P E F = 2 3

The probability of solving specific problem independently by A , B and C are 1 2 , 1 3 and 1 4 respectively. If all try to solve the problem independently, find the probability that the problem is solved.

Probability that A is able to solve = P ( A ) = 1 2 ⇒ P A ' = 1 2 Probability that B is able to solve = P ( B ) = 1 3 ⇒ P B ' = 2 3 Probability that C is able to solve = P ( C ) = 1 4 ⇒ P C ' = 3 4 P the problem is solved = 1 - P A ' P B ' P C ' = 1 - 1 2 × 2 3 × 3 4 = 1 - 1 4 = 3 4

A dice is thrown 6 times. If getting an even number is success, find probability of getting 4 successes.

Given n = 6 S = 1 , 2 , 3 , 4 , 5 , 6 = 6 Probability of getting an even number p = 3 6 = 1 2 q = 1 - p = 1 - 1 2 = 1 2 Binomial probability distribution is given by P X = r = C r n p r q n - r ; r = 0 , 1 , 2 . . . . P ( X = 4 ) = C 4 6 1 2 4 1 2 6 - 4 C 4 = 6 1 2 6 = 15 64 ∴ Probability of getting 4 successes = 15 64 .

A random variable X has the following probability distribution: X 0 1 2 3 4 5 6 7 P X 0 k 2 k 2 k 3 k k 2 2 k 2 7 k 2 + k Find k .

Given that, X 0 1 2 3 4 5 6 7 P X 0 k 2 k 2 k 3 k k 2 2 k 2 7 k 2 + k Since the sum of all the probability in a probability distribution is always unity. ∴ P ( X = 0 ) + P ( x = 1 ) + . . . . . . + P ( x = 7 ) = 1 = 0 + k + 2 k + 2 k + 3 k + k 2 + 2 k 2 + 7 k 2 + k = 1 = 10 k 2 + 9 k - 1 = 0 ⇒ ( 10 k - 1 ) ( k + 1 ) = 0 ⇒ k ≠ - 1 ∵ k ≥ 0 ⇒ 10 k - 1 = 0 ⇒ k = 1 10