Theory of Expectation :: Drawing Balls, Cards, Items, Products

Solution

"x" indicate the number of red marbles drawn [Since you are required to find E (x), the variable would represent the number of red marbles drawn] The number of red marbles that can be drawn would be 0 when "0 Red and 3 White" marbles are drawn 1 when "1 Red and 2 White" marbles are drawn 2 when "2 Red and 1 White" marbles are drawn 3 when "3 Red and 0 White" marbles are drawn ⇒ The values carried by the variable ("x") would be either 0, 1, 2 or 3 ⇒ "X" is a discrete random variable with range = {0, 1, 2, 3} "X" represents the random variable and P(X = x) represents the probability that the value within the range of the random variable is a specified value of "x" Total number of marbles = 8 Red + 6 White = 14 Total Number of possible choices (in drawing the three marbles) = Number of ways in which the three marbles can be drawn from the total 14 = 14C3 ⇒ n = 14 × 13 × 12 3 × 2 × 1 = 364 Probabilty that the three marbles drawn would be "0 Red and 3 White" ⇒ P(0R3W) = 8C0 × 6C3 14C3 Red White Total Available 8 6 14 To Choose 0 3 3 Choices 8C0 6C3 14C3 = 1 × 6 × 5 × 4 3 × 2 × 1 364 = 20 364 = 5 91 "1 Red and 2 White" ⇒ P(1R2W) = 8C1 × 6C2 14C3 Red White Total Available 8 6 14 To Choose 1 2 3 Choices 8C1 6C2 14C3 = 8 1 × 6 × 5 2 × 1 364 = 8 × 15 364 = 30 91 "2 Red and 1 White" ⇒ P(2R1W) = 8C2 × 6C1 14C3 Red White Total Available 8 6 14 To Choose 2 1 3 Choices 8C2 6C1 14C3 = 8 × 7 2 × 1 × 6 1 364 = 28 × 6 364 = 42 91 "3 Red and 0 White" ⇒ P(3R0W) = 8C3 × 6C0 14C3 Red White Total Available 8 6 14 To Choose 3 0 3 Choices 8C3 6C0 14C3 = 8 × 7 × 6 3 × 2 × 1 × 1 364 = 56 364 = 14 91 Probability for the number of red balls to be 0 ⇒ P(X = 0) = P(0R3W) = 5 91 1 ⇒ P(X = 1) = P(1R2W) = 30 91 2 ⇒ P(X = 2) = P(2R1W) = 42 91 3 ⇒ P(X = 3) = P(3R0W) = 14 91 The probabilty distribution of "x" would be x 0 1 2 3 P(X = x) 5 91 30 91 42 91 14 91 Calculations for Mean and Standard Deviations x P (X = x) px [x × P (X = x)] x2 px2 [x2 × P (X = x)] 0 5 91 0 0 0 1 30 91 30 91 1 30 91 2 42 91 84 91 4 168 91 3 14 91 42 91 9 126 91 Total 1 156 91 324 91 = 1.7143 = 3.56 The expected number of red marbles drawn ⇒ Expectation of "x" ⇒ E (x) = Σ px = 1.7143 Variance of the number of red marbles drawn ⇒ var (x) = E (x2) − (E(x))2 ⇒ var (x) = Σ px2 − (Σ px)2 = 3.56 − (1.7143)2 = 3.56 − 2.94 = 0.62 Standard Deviation of the number of red balls drawn ⇒ SD (x) = + √ Var (x) = + √ 0.62 = + 0.79