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.
Answer
363.9k+ views
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.
Last updated date: 29th Sep 2023
•
Total views: 363.9k
•
Views today: 6.63k
Recently Updated Pages
What is the Full Form of DNA and RNA

What are the Difference Between Acute and Chronic Disease

Difference Between Communicable and Non-Communicable

What is Nutrition Explain Diff Type of Nutrition ?

What is the Function of Digestive Enzymes

What is the Full Form of 1.DPT 2.DDT 3.BCG

Trending doubts
How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Briefly mention the contribution of TH Morgan in g class 12 biology CBSE

The equation xxx + 2 is satisfied when x is equal to class 10 maths CBSE

How many millions make a billion class 6 maths CBSE

Which are the Top 10 Largest Countries of the World?

Number of Prime between 1 to 100 is class 6 maths CBSE

One cusec is equal to how many liters class 8 maths CBSE

How many crores make 10 million class 7 maths CBSE
