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Probability

Question
CBSEENMA12033821

Two numbers are selected at random (without replacement) from the first six positive integers. Let X denote the larger of the two numbers obtained. Find E(X).

Solution

Since 1 cannot be greater than the other selected number.
∴    X can take values 2, 3, 4, 5, 6.
∴    P(X = 2) = P(2 and a number less than 2 are selected)
= P(1 and 2 are selected)
Syntax error from line 1 column 349 to line 1 column 452. Unexpected '<mlongdiv '.
P(X = 3) = P(3 and a number less than 3 are selected)
      equals space fraction numerator 2 over denominator straight C presuperscript 6 subscript 2 end fraction space equals 2 over 15
[∵ (1, 3) and (2, 3) can be selected]
P(X = 4) = P(4 and a number less than 4 are selected)
equals space fraction numerator 3 over denominator straight C presuperscript 6 subscript 2 end fraction space equals space 3 over 15
[∵ (1, 4), (2, 4) and (3, 4) can be selected]
P(X = 5) = P(5 and a number less than 5 are selected)
equals space fraction numerator 4 over denominator straight C presuperscript 6 subscript 2 end fraction space equals 4 over 15.
[∵ (1, 5), (2, 5), (3, 5) and (4, 5) can be selected]
P(X = 6) = P(6 and a number less than 6 are selected)
equals space fraction numerator 5 over denominator straight C presuperscript 6 subscript 2 end fraction space equals space 5 over 15
[∵ (1, 6), (2, 6), (3, 6), (4, 6) and (5, 6) can be selected]
∴ probability distribution table is

therefore  Expectation of X = E(X)  = sum from blank to blank of cross times space straight P left parenthesis straight X right parenthesis
                          equals space 2 cross times 1 over 15 plus 2 space cross times space 2 over 15 plus 4 space cross times 3 over 15 plus 5 space cross times 4 over 15 plus 6 space cross times space 5 over 15
equals space fraction numerator 2 plus 6 plus 12 plus 20 plus 30 over denominator 15 end fraction space equals 70 over 15 space equals 14 over 3.

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