Answer to Question #91727 in Statistics and Probability for Sibel Gercek

Question

(3.26) From a box containing 4 black balls and 2 green balls, 3 balls are drawn in succession, each ball being replaced in the box before the next draw is made.
a) Find the Probability Distribution for X, the number of green balls. Explain your answer.
b) Calculate the expected number of green balls.

Answer 1

Answer

a) {8/27, 4/9, 2/9, 1/27}

b) 1

Explanation

a) Because each ball is replaced in the box before the next draw is made, the number of green balls X is a discrete random variable, which can take four possible values: 0, 1, 2, 3. Let us find the probabilities that X takes one of these values. We denote by A the event, that we draw a green ball from the box in one trial. The number of all possible outcomes in one trial of the experiment n = 6. For this event the number of favourable outcomes m = 2. So with the help of classical definition of probability the probability of this event is found as p = p(A) = m/n = 2/6 = 1/3. Because each ball is replaced in the box before the next draw is made, we have a sequence of three Bernoulli trials with success probability p = 1/3 and failure probability q = 1 – p = 2/3.

The probability p_n(k) that exactly k successes in n trials will occur, is calculated by Bernoulli formula p_n(k) = C(n,k)p^kq^n-k, where the number of combinations of n different elements taken k at a time C(n,k) = n!/k!/(n-k)!. In case of our problem n = 3, k takes one of four values 0, 1, 2, 3. So we have, p(X=0) = p₃(0) = C(3,0)(1/3)⁰(2/3)³ = 3!/0!/3!*8/27 = 8/27; p(X=1) = p₃(1) = C(3,1)(1/3)¹(2/3)² = 3!/1!/2!*1/3*4/9 = 3*4/27 = 4/9; p(X=2) = p₃(2) = C(3,2)(1/3)²(2/3)¹ = 3!/2!/1!*1/9*2/3 = 3*1/9*2/3 = 2/9; p(X=3) = p₃(3) = C(3,3)(1/3)³(2/3)⁰ = 3!/3!/0!*1/9*2/3 = 1*1/27*1 = 1/27. Hence, the set {8/27, 4/9, 2/9, 1/27} forms the distribution of X. Obviously, the discrete random variable X is completely defined by the table:

x_i    0    1    2    3

p_i   8/27   4/9   2/9   1/27

b) The quantity E(X) = x₁*p₁ + x₂*p₂ + x_n*p_n is referred to as the expectation (the mean value or the first moment) of the random variable. In case of our problem E(X) = 0*8/27+1*4/9+2*2/9+3*1/27 = 1.