Let X ~ B 10 , 0 · 2 . Find P X = 1 .

A binomial distribution with n -Bernoulli trials and probability of success in each trial as p is denoted by B ( n , p ) or X ~ B ( n , p ) . Given that X ~ B 10 , 0 . 2 , to find P X = 1 . The probability of x successes P X = x is also denoted by P ( x ) and is given by P ( x ) = C x n q n - x p x , x = 0 , 1 , . . . , n , q = 1 - p Here, n = 10 , p = 0 . 2 , q = 1 - 0 . 2 = 0 . 8 Now, P X = 1 = P 1 = C 1 10 0 . 8 10 - 1 0 . 2 1 ⇒ P ( 1 ) = 10 1 × 0 . 8 10 - 1 0 . 2 1 ⇒ P ( 1 ) = 10 × 0 . 8 9 × 2 10 = 2 × 0 . 8 9 Therefore, P ( X = 1 ) = 2 × 0 . 8 9

Let X ~ B 10 , 0 · 2 . Find P X ≥ 1 .

A binomial distribution with n -Bernoulli trials and probability of success in each trial as p is denoted by B ( n , p ) or X ~ B ( n , p ) . Given that X ~ B 10 , 0 . 2 , to find P X ≥ 1 . The probability of x successes P X = x is also denoted by P ( x ) and is given by P ( x ) = C x n q n - x p x , x = 0 , 1 , . . . , n , q = 1 - p Here, n = 10 , p = 0 . 2 , q = 1 - 0 . 2 = 0 . 8 Now, P X ≥ 1 = 1 - P X = 0 = 1 - C 0 10 0 . 8 10 - 0 0 . 2 0 ⇒ P ( 0 ) = 1 - 1 × 0 . 8 10 × 1 ⇒ P ( 0 ) = 1 - 0 . 8 10 Therefore, P ( X ≥ 1 ) = 1 - 0 . 8 10

Let X ~ B 10 , 0 · 2 , Find P X ≤ 8 .

A binomial distribution with n -Bernoulli trials and probability of success in each trial as p is denoted by B ( n , p ) or X ~ B ( n , p ) . Given that X ~ B 10 , 0 . 2 , to find P X ≤ 8 . The probability of x successes P X = x is also denoted by P ( x ) and is given by P ( x ) = C x n q n - x p x , x = 0 , 1 , . . . , n , q = 1 - p Here, n = 10 , p = 0 . 2 , q = 1 - 0 . 2 = 0 . 8 Now, P X ≤ 8 = 1 - P X > 8 1 - P X > 8 = 1 - P X = 9 + P X = 10 = 1 - C 9 10 0 . 8 1 0 . 2 9 + C 10 10 0 . 8 0 0 . 2 10 = 1 - 10 × 0 . 8 × 0 . 2 9 + 1 × 0 . 2 10 = 1 - 0 . 2 9 8 + 0 . 2 = 1 - 8 . 2 × 0 . 2 9 Therefore, P X ≤ 8 = 1 - 8 . 2 × 0 . 2 9

Let X ~ B n , p If E X = 5 and Var X = 2 · 5 , find n and p

Let X ~ B n , p then mean or expected value of random variable X is denoted by E ( X ) and the variance is denoted by V a r X . E X = n p and V a r X = n p q n = number of trials, p = probability of success and q = probability of failure We have E X = 5 , V X = 2 . 5 , to find n and p . Now, n p q n p = 2 . 5 5 ⇒ q = 0 . 5 and p = 1 - 0 . 5 = 0 . 5 n p = 5 ⇒ n × 0 . 5 = 5 and n = 5 0 . 5 = 10 Therefore, n = 10 and p = 0 . 5 .

Five cards are drawn successively with replacement from a well shuffled deck of 52 cards. What is the probability that all the five cards are spades?

Let X represent the number of spade cards among the five cards drawn. Since the drawing of card is with replacement, the trials are Bernoulli trials. In a well shuffled deck of 52 cards, there are 13 spade cards. Probability of drawing a spade = p = 13 52 = 1 4 Probability of drawn card is not spade = q = 1 - p = 1 - 1 4 = 3 4 Probability that all the five cards are spades = P X = 5 = C 5 5 1 4 5 3 4 5 - 5 ∵ P X = r = C r n p r q n - r = 1 4 5 Which is the required probability.

A person buys a lottery ticket in 50 lotteries, in each of which his chance of winning a prize is 1 100 . What is the probability that he will win a prize atleast once.

Given that a person buys a lottery ticket in 50 lotteries Let X denote the number of times the person wins the lottery. Then, X follows a binomial distribution with n = 50 . Let p be the probability of winning a prize ∴ p = 1 100 , q = 1 - 1 100 = 99 100 According to binomial probability distribution, P X = r = C r n q n - r p r P winning atleast once = P ( X ≥ 1 ) = 1 - P X = 0 = 1 - 50 C 0 1 100 0 99 100 50 - 0 = 1 - 99 100 50