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June 2012 p62 q2
2941
The random variable X has the probability distribution shown in the table.
x
2
4
6
P(X = x)
0.5
0.4
0.1
Two independent values of X are chosen at random. The random variable Y takes the value 0 if the two values of X are the same. Otherwise the value of Y is the larger value of X minus the smaller value of X.
Draw up the probability distribution table for Y.
Find the expected value of Y.
Solution
(i) To find the probability distribution of Y, consider the possible outcomes when two values of X are chosen:
Y = 0: This occurs when both values are the same. The probabilities are:
P(2, 2) = 0.5 × 0.5 = 0.25
P(4, 4) = 0.4 × 0.4 = 0.16
P(6, 6) = 0.1 × 0.1 = 0.01
Total probability for Y = 0 is 0.25 + 0.16 + 0.01 = 0.42.
Y = 2: This occurs when the values are (2, 4), (4, 2), (4, 6), or (6, 4). The probabilities are:
P(2, 4) = 0.5 × 0.4 = 0.2
P(4, 2) = 0.4 × 0.5 = 0.2
P(4, 6) = 0.4 × 0.1 = 0.04
P(6, 4) = 0.1 × 0.4 = 0.04
Total probability for Y = 2 is 0.2 + 0.2 + 0.04 + 0.04 = 0.48.
Y = 4: This occurs when the values are (2, 6) or (6, 2). The probabilities are:
