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Probability

Question

Wired Faculty App

Suppose that two cards are drawn at random from a deck of cards. Let X be the number of aces obtained. Then the value of E(X) is
$37 over 221$
$5 over 13$
$1 over 13$
$2 over 13$

Solution

X is the number of aces obtained. So, X can take values 0, 1, 2.

straight P left parenthesis straight X space equals space 0 right parenthesis space equals space straight P left parenthesis no space ace space is space drawn right parenthesis space equals space fraction numerator straight C presuperscript 48 subscript 2 over denominator straight C presuperscript 52 subscript 2 end fraction space equals fraction numerator 48 space cross times space 47 over denominator 52 space cross times space 51 end fraction space equals space 188 over 221 straight P left parenthesis straight X space equals space 1 right parenthesis space equals space straight P left parenthesis one space ace space and space non minus ace right parenthesis space equals space fraction numerator straight C presuperscript 4 subscript 1 cross times straight C presuperscript 48 subscript 1 over denominator straight C presuperscript 52 subscript 2 end fraction space equals space fraction numerator 4 space cross times space 48 over denominator begin display style fraction numerator 52 space cross times 51 over denominator 1 space cross times 2 end fraction end style end fraction space space space space space space space space space space space space space space space space space equals space fraction numerator 4 space cross times space 48 space cross times space 1 space cross times space 2 over denominator 52 space cross times space 51 end fraction space equals space 32 over 221

and

straight P left parenthesis straight X space equals space 2 right parenthesis space equals space straight P left parenthesis two space aces space are space drawn right parenthesis space equals space fraction numerator straight C presuperscript 4 subscript 2 over denominator straight C presuperscript 52 subscript 2 end fraction space equals fraction numerator 4 space cross times space 3 over denominator 52 space cross times space 51 end fraction space equals space 1 over 221

∴ probability distribution of X is

therefore space space space straight E left parenthesis straight X right parenthesis space equals space sum straight x space straight P left parenthesis straight X right parenthesis space equals space 0 space cross times 188 over 221 plus 1 space cross times space 32 over 221 plus 2 space cross times 1 over 221 space space space space space space space space space space space space space space space space equals space 0 plus 32 over 221 plus 2 over 221 space equals space 34 over 221 space equals 2 over 13.

∴ (D) is correct answer.

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