15. A small deck of cards contains five red cards, four blue cards and one green

Question

15. A small deck of cards contains five red cards, four blue cards and one green card. We will shuffle the deck and select three cards without replacement. Let X be the number of blue cards that are selected. The probability distribution of X is shown below: P(X- 0.1670.500 0.300 0.033 The expected value of X is calculated to be E(X) = 1.2. What is the variance of X? (A) 0.56 (B) 0.61 (C) 0.68 (D) 0.72 (E) 0.85 16. What would be the variance of X if we had instead selected the three cards with replacement? (A) 0.56 (B) 0.61 (C) 0.68 (D) 0.72 (E) 0.85

Explanation / Answer

15) E(X) = 1.2

E(X2) = 02 * 0.167 + 12 * 0.5 + 22 * 0.3 + 32 * 0.033 = 1.997

Var(X) = E(X2) - (E(X))2 = 1.997 - 1.22 = 0.557

Option-A) 0.56

16) If we select with replacement then this is a binomial distribution.

p = P(blue card) = 4/10 = 0.4

n = 3

Var = n * p * (1 - p) = 3 * 0.4 * 0.6 = 0.72

Option-D) 0.72

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