For the following probability distribution of X . Find k X 0 1 2 3 4 P ( X = x ) k 2 k 4 k 2 k k

X 0 1 2 3 4 P ( X = x ) k 2 k 4 k 2 k k Given, Sum of all the probability = 1 k + 2 k + 4 k + 2 k + k = 1 ∴ k = 1 10

For the following probability distribution of X X 0 1 2 3 4 P ( X = x ) k 2 k 4 k 2 k k Find P ( X > 2 ) , P ( X < 3 ) , P ( X < 1 ) .

X 0 1 2 3 4 P ( X = x ) k 2 k 4 k 2 k k Given, As we know, total probability = 1 So, k + 2 k + 4 k + 2 k + k = 1 ⇒ 10 k = 1 ⇒ k = 1 10 Now, p ( x > 2 ) = p ( 3 ) + p ( 4 ) = 3 k = 3 10 And p ( x < 3 ) = p ( 1 ) + p ( 2 ) + p ( 0 ) = 2 k + 4 k + k = 7 k = 7 10 And p ( x < 1 ) = p ( 0 ) = k = 1 10

A random variable X has the following probability distribution : X 0 1 2 3 4 5 6 P X C 2 C 2 C 3 C C 2 2 C 2 7 C 2 + C Find the value of C and also calculate the mean of the distribution.

We will find the value of C and also calculate the mean of the distribution. A random variable X has the following probability distribution : X 0 1 2 3 4 5 6 P X C 2 C 2 C 3 C C 2 2 C 2 7 C 2 + C We know that sum of all probabilities of probability distribution be equal to 1 . So, C + 2 C + 2 C + 3 C + C 2 + 2 C 2 + 7 C 2 + C = 1 ⇒ 10 C 2 + 9 C - 1 = 0 ⇒ 10 C - 1 C + 1 = 0 ⇒ C = - 1 , 1 10 If C = - 1 , P 0 = - 1 which is not possible. ∴ C = 1 10 We know that Mean = μ = E X = ∑ x i P x i ∴ μ = 0 × C + 1 × 2 C + 2 × 2 C + 3 × 3 C + 4 × C 2 + 5 × 2 C 2 + 6 7 C 2 + C = 56 C 2 + 21 C = 56 × 1 100 + 21 10 ∵ C = 1 10 = 56 + 210 100 = 266 100 = 133 50 Therefore, 1 10 is the value of C and 133 50 is the mean of the distribution.

Following is the distribution function F (x) of a discrete r.v. X x 1 2 3 4 5 6 F ( x ) 0 . 2 0 . 37 0 . 48 0 . 62 0 . 85 1 Find P ( X ≤ 3 ) , P ( 2 < X < 5 ) .

x 1 2 3 4 5 6 f ( x ) 0 . 2 0 . 37 0 . 48 0 . 62 0 . 85 1 P ( x ) 0 . 2 0 . 17 0 . 11 0 . 14 0 . 23 0 . 15 P ( X ≤ 3 ) = P ( x = 1 ) + P ( x = 2 ) + P x = 3 = 0 . 2 + 0 . 17 + 0 . 11 = 0 . 30 Now, P ( 2 < X < 5 ) = P ( x = 3 ) + P ( x = 4 ) ⇒ P ( 2 < X < 5 ) = 0 . 11 + 0 . 14 Hence, P ( 2 < X < 5 ) = 0 . 25

Two biased dice are thrown together. For the first die P ( 6 ) = 1 2 , the other scores being equally likely while for the second die, P ( 1 ) = 2 5 and the other scores are equally likely. Find the probability distribution of " the number of ones seen ".

For the first die, P ( 6 ) = 1 2 ⇒ P ( 1 ) + P ( 2 ) + P ( 3 ) + P ( 4 ) + P ( 5 ) + 1 2 = 1 ⇒ 5 P ( 1 ) = 1 2 [ ∵ P ( 1 ) = P ( 2 ) = P ( 3 ) = P ( 4 ) = P ( 5 ) ] ⇒ P ( 1 ) = 1 10 and P not 1 = 9 10 For the second die, P ( 1 ) = 2 5 ⇒ P ( not 1 ) = 1 - 2 5 = 3 5 Let X = Number of ones seen when two dice are thrown together. P ( X = 0 ) = P ( not 1 ) · P ( not 1 ) = 9 10 · 3 5 = 27 50 = 0 . 54 P ( X = 1 ) = P ( not 1 ) · P ( 1 ) + P ( 1 ) · P ( not 1 ) = 9 10 · 2 5 + 1 10 · 3 5 = 21 50 = 0 . 42 P ( X = 2 ) = P ( 1 ) · P ( 1 ) = 1 10 · 2 5 = 2 50 = 0 . 04 Hence, the required probability distribution is as below. X 0 1 2 P ( X ) 0 . 54 0 . 42 0 . 04