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  • 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.

Q. 12    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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Let   A : Event of choosing a diamond card.

        B :  Event of not choosing a diamond card.

P(A)=\frac{13}{52}=\frac{1}{4}

P(B)=\frac{39}{52}=\frac{3}{4}

X : The lost card.

If lost card is diamond then 12 diamond cards are left out of 51 cards.

Two diamond cards are drawn out of 12 diamond cards in ^{12}\textrm{C}_2 ways.

Similarly, two cards are drawn out of 51 cards in ^{51}\textrm{C}_2 ways.

Probablity of getting two diamond cards when one diamond is lost : P(X|A)= \frac{^{12}\textrm{C}_2}{^{51}\textrm{C}_2}

                                                                                                         P(X|A)=\frac{12!}{10!\times 2!}\times \frac{49!\times 2!}{51!}

                                                                                                          P(X|A)=\frac{11\times 12}{50\times 51}

                                                                                                            P(X|A)=\frac{22}{425}

If lost card is not diamond then 13 diamond cards are left out of 51 cards.

Two diamond cards are drawn out of 13 diamond cards in ^{13}\textrm{C}_2 ways.

Similarly, two cards are drawn out of 51 cards in ^{51}\textrm{C}_2 ways.

Probablity of getting two diamond cards when one diamond is not lost : P(X|B)= \frac{^{13}\textrm{C}_2}{^{51}\textrm{C}_2}

                                                                                                         P(X|B)=\frac{13!}{11!\times 2!}\times \frac{49!\times 2!}{51!}

                                                                                                          P(X|B)=\frac{13\times 12}{50\times 51}

                                                                                                            P(X|B)=\frac{26}{425}

The probability of the lost card being a diamond : P(B|X)

P(B|X)= \frac{P(B).P(X|B)}{P(B).P(X|B)+P(A).P(X|A)}

P(B|X)= \frac{\frac{1}{4}\times \frac{22}{425}}{\frac{1}{4}\times \frac{22}{425}+\frac{3}{4}\times \frac{26}{425}}

P(B|X)= \frac{\frac{11}{2}}{25}

P(B|X)= \frac{11}{50}

Hence, the probability of the lost card being a diamond : 

                                                                                          P(B|X)= \frac{11}{50}

Posted by

seema garhwal

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