An urn contains 4 red balls, 3 blue balls, 2 green balls and one black ball. One

Question

An urn contains 4 red balls, 3 blue balls, 2 green balls and one black ball. One ball is selected at random. Consider games X defined as follows.

(a) Find the probabilities P(X > 1.00) and P(X 1.00)

. (b) Find the expected value, E(X).

(c) Find the probability standard deviation, (X).

(d) Find the cumulative distribution functions F(x) = P(X x).

2. From the urn of Problem 1, ten balls are selected at random with replacement.

(a) Estimate the probability that four or less red balls are selected.

(b) Estimate the probability that more than four blue balls are selected.

Explanation / Answer

1 a) P(x>1) = P(green) + P(black) = 2/10 + 1/10 = 3/10

P(x>=1) = 3/10 + p(blue) = 3/10 + 3/10 = 2/5

E(x) = sum of (X x P(X)) = (-5x5/10) + (1x3/10) + (2.5x2/10) + (10x1/10)

= -7/10 = -0.7

C) variance = (-5+0.7)2x0.5 + (1+0.7)2x0.3+(2.5+0.7)2x0.2 + (10+0.7)2 x 0.1

= 23.61

Standard deviation = square root (23.61) = 4.86

D) x is not a number, so, please define what cumulative function is needed here

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