Question

A card from a pack of 52 cards is lost. From the remaining cards of the pack, two cards are drawn and are found to be both diamonds. Find the probability of the lost card being a diamond.

Solution

Let E₁, E₂, E he the events
E₁ : ‘lost card is a diamond’,
E₂: ‘lost card is not a diamond’,
E : ‘two cards drawn from the remaining pack are diamonds’
$therefore space space space straight P left parenthesis straight E subscript 1 right parenthesis space equals space 13 over 52 space equals space 1 fourth comma space space space straight P left parenthesis straight E subscript 2 right parenthesis space equals space 39 over 52 space equals space 3 over 4$
Also,    $straight P left parenthesis straight E space vertical line thin space straight E subscript 1 right parenthesis space equals space fraction numerator straight C presuperscript 12 subscript 2 over denominator straight C presuperscript 51 subscript 2 end fraction space equals fraction numerator begin display style fraction numerator 12 space cross times space 11 over denominator 1 space cross times 2 end fraction end style over denominator begin display style fraction numerator 51 space cross times 50 over denominator 1 space cross times 2 end fraction end style end fraction space equals space fraction numerator 12 space cross times space 11 over denominator 51 space cross times space 50 end fraction$
and    $straight P left parenthesis straight E vertical line straight E subscript 2 right parenthesis space equals space fraction numerator straight C presuperscript 13 subscript 2 over denominator straight C presuperscript 51 subscript 2 end fraction space equals space fraction numerator 13 space cross times space 12 over denominator 51 space cross times space 50 end fraction space$
Required probability = $straight P left parenthesis straight E subscript 1 space vertical line thin space straight E right parenthesis$
   $equals space fraction numerator straight P left parenthesis straight E subscript 1 right parenthesis thin space straight P left parenthesis straight E space vertical line thin space straight E subscript 1 right parenthesis over denominator straight P left parenthesis straight E subscript 1 right parenthesis thin space straight P left parenthesis straight E vertical line straight E subscript 1 right parenthesis space plus space straight P left parenthesis straight E subscript 2 right parenthesis thin space straight P left parenthesis straight E vertical line straight E subscript 2 right parenthesis end fraction equals space fraction numerator begin display style 1 fourth end style cross times begin display style fraction numerator 12 cross times 11 over denominator 51 cross times 50 end fraction end style over denominator begin display style 1 fourth end style cross times begin display style fraction numerator 12 cross times 11 over denominator 51 cross times 50 end fraction end style plus begin display style 3 over 4 end style cross times begin display style fraction numerator 13 space cross times 12 over denominator 51 cross times 50 end fraction end style end fraction space equals fraction numerator 12 space cross times space 11 over denominator 12 space cross times 11 space plus space 13 space cross times 12 space cross times 3 end fraction equals space fraction numerator 132 over denominator 132 plus 468 end fraction space equals 132 over 600 equals space 11 over 50.$

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Probability

A card from a pack of 52 cards is lost. From the remaining cards of the pack, two cards are drawn and are found to be both diamonds. Find the probability of the lost card being a diamond.

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