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# A pack of playing cards was found to contain only 51 cards. If the first 13 cards which are examined are all red, what is the probability that the missing card is black? Verified
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Hint: To solve the question, we have to apply Bayes theorem of conditional probability and calculate the required probabilities to calculate the probability of the missing card is black.

Let A be the event of the missing card is black and B be the event of the missing card is red.
Let C be the event of, the first 13 cards are examined in red colour.

The probability of the missing card is black can be calculated using Bayes theorem, which is given by the formula $P\left( {A}/{C}\; \right)=\dfrac{P\left( A \right)P\left( {C}/{A}\; \right)}{P\left( A \right)P\left( {C}/{A}\; \right)+P\left( B \right)P\left( {C}/{B}\; \right)}$. ……(1)
The given number of cards found in a pack of playing card = 51

Generally, the number of cards present in a pack of playing card = 52

Where the 52 cards generally are distributed into 26 black cards and 26 red cards.

$\Rightarrow$ The missing card can be either black or red.

The probability of the event A is equal to $\dfrac{1}{2}$

$\Rightarrow P\left( A \right)=\dfrac{1}{2}$

The probability of the event B is equal to $\dfrac{1}{2}$

$\Rightarrow P\left( B \right)=\dfrac{1}{2}$

The probability of selecting the 13 red cards given that the missing card is black $P\left( {C}/{A}\; \right)$= Ratio of selecting 13 examined red cards from the 26 red cards of the pack of cards to selecting 13 red cards from the given 51 cards.

$P\left( {C}/{A}\; \right)=\dfrac{{}^{26}{{C}_{13}}}{{}^{51}{{C}_{13}}}$

The probability of selecting the 13 red cards given that the missing card is red $P\left( {C}/{B}\; \right)$= Ratio of selecting 13 examined red cards from the 25 red cards of the pack of cards to selecting 13 red cards from the given 51 cards.

$P\left( {C}/{B}\; \right)=\dfrac{{}^{25}{{C}_{13}}}{{}^{51}{{C}_{13}}}$

By substituting the above values in the equation (1) we get,

$P\left( {A}/{C}\; \right)=\dfrac{\dfrac{1}{2}\times \dfrac{{}^{26}{{C}_{13}}}{{}^{51}{{C}_{13}}}}{\dfrac{1}{2}\times \dfrac{{}^{26}{{C}_{13}}}{{}^{51}{{C}_{13}}}+\dfrac{1}{2}\times \dfrac{{}^{25}{{C}_{13}}}{{}^{51}{{C}_{13}}}}$

$=\dfrac{\dfrac{{}^{26}{{C}_{13}}}{{}^{51}{{C}_{13}}}}{\dfrac{{}^{26}{{C}_{13}}}{{}^{51}{{C}_{13}}}+\dfrac{{}^{25}{{C}_{13}}}{{}^{51}{{C}_{13}}}}$

$\Rightarrow P({A}/{C}\;)=\dfrac{{}^{26}{{C}_{13}}}{{}^{26}{{C}_{13}}+{}^{25}{{C}_{13}}}$

We know that ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ and $n!=n\left( n-1 \right)\left( n-2 \right).....2\times 1$

$\Rightarrow P({A}/{C}\;)=\dfrac{\dfrac{26!}{13!\left( 26-13 \right)!}}{\dfrac{26!}{13!\left( 26-13 \right)!}+\dfrac{25!}{13!\left( 25-13 \right)!}}$

$=\dfrac{\dfrac{26!}{13!13!}}{\dfrac{26!}{13!13!}+\dfrac{25!}{13!12!}}$
$=\dfrac{\dfrac{26!}{13!}}{\dfrac{26!}{13!}+\dfrac{25!}{12!}}$

We know that $n!=n\left( n-1 \right)!$

$=\dfrac{\dfrac{26\times 25!}{13\times 12!}}{\dfrac{26\times 25!}{13\times 12!}+\dfrac{25!}{12!}}$

$=\dfrac{\dfrac{25!}{12!}\times \left( \dfrac{26}{13} \right)}{\dfrac{25!}{12!}\times \left( \dfrac{26}{13}+1 \right)}$

$=\dfrac{\left( \dfrac{26}{13} \right)}{\left( \dfrac{26}{13}+1 \right)}$

$=\dfrac{\left( \dfrac{26}{13} \right)}{\left( \dfrac{26+13}{13} \right)}$

$=\dfrac{26}{\left( 26+13 \right)}=\dfrac{26}{39}=\dfrac{2}{3}$

$\Rightarrow P({A}/{C}\;)=\dfrac{2}{3}$

Thus, the probability of the missing card is black is equal to $\dfrac{2}{3}$

Note: The possibility of mistake can be not applying Bayes rule which is an important step to calculate the probability of the missing card is black. The other possible mistake can be the calculation mistake since the calculations involve multiple fractions and factorial calculations.
Last updated date: 22nd Sep 2023
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