# 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 diamonds .Find the probability of the lost card being a diamond.

Last updated date: 23rd Mar 2023

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Answer

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Hint: Here, we will have to find the probability of the lost card being a diamond. So, we use the Bayes theorem since it depends on the prior events.

As we know in a pack of 52 cards, 13 cards are diamond.

Let,

\[{E_1}\]: Event that lost card is diamond

\[{E_2}\]: Event that lost card is not diamond

A: Events where two cards are drawn are diamond.

Now,

$P({E_1})$=Probability that lost card is diamond =$\frac{{13}}{{52}} = \frac{1}{4}$

Similarly, the probability that the lost card is not diamond is

$P({E_2})$=Probability that lost card is not diamond =\[1 - P({E_1}) = 1 - \frac{1}{4} = \frac{3}{4}\].

Now let us find the probability of getting 2 diamond cards if the lost card is diamond. i.e..,

$

P(A/{E_1}) = \frac{{{\text{selection of two diamond cards from 12(13 - 1) diamond cards}}}}{{{\text{Selection of any two cards from 51 cards}}}} \\

P(A/{E_1}) = \frac{{{}^{12}{C_2}}}{{{}^{51}{C_2}}} = \frac{{12 \times 11}}{{51 \times 50}} = \frac{{22}}{{425}} \\

$

Similarly let us find the probability of getting 2 diamond cards if the lost card is not diamond.

$

P(A/{E_2}) = \frac{{{\text{selection of two diamond cards form 13 diamond cards}}}}{{{\text{Selection of any two cards from 51 cards}}}} \\

P(A/{E_2}) = \frac{{{}^{13}{C_2}}}{{{}^{51}{C_2}}} = \frac{{13 \times 12}}{{51 \times 50}} = \frac{{26}}{{425}} \\

$

Now, the probability of lost card being a diamond if two cards drawn are found to be both diamond .i.e..,

$P({E_1}/A) = \frac{{P({E_1}).P(A/{E_1})}}{{P({E_1}).P(A/{E_1}) + P({E_2}).P(A/{E_2})}} \to (1)$

Substituting the values in the equation (1) we get

\[P({E_1}/A) = \frac{{\frac{1}{4} \times \frac{{22}}{{425}}}}{{\frac{1}{4} \times \frac{{22}}{{425}} + \frac{3}{4} \times \frac{{26}}{{425}}}} = \frac{{11}}{{50}}\]

Therefore, the required probability is$\frac{{11}}{{50}}$.

Note: In these types of problems, we need to consider all the possible outcomes for an event and use the Bayes theorem to solve.

As we know in a pack of 52 cards, 13 cards are diamond.

Let,

\[{E_1}\]: Event that lost card is diamond

\[{E_2}\]: Event that lost card is not diamond

A: Events where two cards are drawn are diamond.

Now,

$P({E_1})$=Probability that lost card is diamond =$\frac{{13}}{{52}} = \frac{1}{4}$

Similarly, the probability that the lost card is not diamond is

$P({E_2})$=Probability that lost card is not diamond =\[1 - P({E_1}) = 1 - \frac{1}{4} = \frac{3}{4}\].

Now let us find the probability of getting 2 diamond cards if the lost card is diamond. i.e..,

$

P(A/{E_1}) = \frac{{{\text{selection of two diamond cards from 12(13 - 1) diamond cards}}}}{{{\text{Selection of any two cards from 51 cards}}}} \\

P(A/{E_1}) = \frac{{{}^{12}{C_2}}}{{{}^{51}{C_2}}} = \frac{{12 \times 11}}{{51 \times 50}} = \frac{{22}}{{425}} \\

$

Similarly let us find the probability of getting 2 diamond cards if the lost card is not diamond.

$

P(A/{E_2}) = \frac{{{\text{selection of two diamond cards form 13 diamond cards}}}}{{{\text{Selection of any two cards from 51 cards}}}} \\

P(A/{E_2}) = \frac{{{}^{13}{C_2}}}{{{}^{51}{C_2}}} = \frac{{13 \times 12}}{{51 \times 50}} = \frac{{26}}{{425}} \\

$

Now, the probability of lost card being a diamond if two cards drawn are found to be both diamond .i.e..,

$P({E_1}/A) = \frac{{P({E_1}).P(A/{E_1})}}{{P({E_1}).P(A/{E_1}) + P({E_2}).P(A/{E_2})}} \to (1)$

Substituting the values in the equation (1) we get

\[P({E_1}/A) = \frac{{\frac{1}{4} \times \frac{{22}}{{425}}}}{{\frac{1}{4} \times \frac{{22}}{{425}} + \frac{3}{4} \times \frac{{26}}{{425}}}} = \frac{{11}}{{50}}\]

Therefore, the required probability is$\frac{{11}}{{50}}$.

Note: In these types of problems, we need to consider all the possible outcomes for an event and use the Bayes theorem to solve.

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