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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.