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Question: A card from a pack of 52 cards is lost from the remaining cards of the pack, two cards are drawn and...

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.

Explanation

Solution

Probability is the branch of mathematics concerning numerical descriptions of how likely an event is to occur, or how likely it is that a proposition is true.
In this first we will apply Bayes theorem to calculate the probability.

Complete answer: As there are 52 cards in a pack of cards and there are 4 types of cards and each has 13 cards.
Let assume E1{E_1}​ be the event for the lost card is Diamond.
As we know:
Given that there are 13 diamonds in the deck.
So, Probability will be P(E1)=favorableoutcomeTotalOutcome=1352=14P\left( {{E_1}} \right) = \dfrac{{favorable\,outcome}}{{Total\,Outcome}} = \dfrac{{13}}{{52}} = \dfrac{1}{4}

Let E1{E_1}​ be the event when the lost card is not a diamond.
Card remaining diamond will be 5213=3952 - 13 = 39
So, P(E2)=favorableoutcomeTotalOutcome=3952=34P\left( {{E_2}} \right) = \dfrac{{favorable\,outcome}}{{Total\,Outcome}} = \dfrac{{39}}{{52}} = \dfrac{3}{4}
Let A be the event that the two cards drawn are both diamonds.
When lost card is diamond
P(AE1)=12C251C2=No. of ways of drawing 2 diamonds cardsTotal Number of way to draw 2 cardP\left( {\dfrac{A}{{{E_1}}}} \right) = \dfrac{{^{12}{C_2}}}{{^{51}{C_2}}} = \dfrac{{{\text{No}}{\text{. of ways of drawing 2 diamonds cards}}}}{{{\text{Total Number of way to draw 2 card}}}}
=661275= \dfrac{{66}}{{1275}}
And when the lost card is not a diamond.
P(AE2)=13C251C2P\left( {\dfrac{A}{{{E_2}}}} \right) = \dfrac{{^{13}{C_2}}}{{^{51}{C_2}}}
=781275= \dfrac{{78}}{{1275}}
Applying Bayes Theorem, where
P(E1A)=P(E1)×P(AE1)P(E1)×P(AE1)+P(E2)×P(AE2)P\left( {\dfrac{{{E_1}}}{A}} \right) = \dfrac{{P\left( {{E_1}} \right) \times P\left( {\dfrac{A}{{{E_1}}}} \right)}}{{P\left( {{E_1}} \right) \times P\left( {\dfrac{A}{{{E_1}}}} \right) + P\left( {{E_2}} \right) \times P\left( {\dfrac{A}{{{E_2}}}} \right)}}
Keeping value in it. We get,
P(E1A)=661275×14661275×14+781275×34=1150P\left( {\dfrac{{{E_1}}}{A}} \right) = \dfrac{{\dfrac{{66}}{{1275}} \times \dfrac{1}{4}}}{{\dfrac{{66}}{{1275}} \times \dfrac{1}{4} + \dfrac{{78}}{{1275}} \times \dfrac{3}{4}}} = \dfrac{{11}}{{50}}
So, the probability of the lost card will be diamond 1150\dfrac{{11}}{{50}} .

Note:
Bayes theorem: The Bayes' theorem essentially describes the probability Total Probability Rule the Total Probability Rule (also known as the law of total probability) is a fundamental rule in statistics relating to conditional and marginal effects of an event based on prior knowledge of the conditions that may be relevant to the event.