Question

A pack of cards is counted with face downwards. It is found that one card is missing. One card is drawn and is found to be red. Find the probability that the missing card is red.

Answer 1

Here, we will use Bays’ theorem to find the required probability. We will find the probability of the missing card to be red and black respectively. Then, we will find the conditional probability of the missing card to be red and black respectively when it is given that the card drawn is red. Then, substituting the values in the formula of Bayes’ Theorem we will get the required probability.

Formula Used:
We will use the following formulas:
1.Bayes’ Theorem: $P\left( {{A_1}|A} \right) = \dfrac{{P\left( {A|{A_1}} \right) \cdot P\left( {{A_1}} \right)}}{{P\left( {A|{A_1}} \right) \cdot P\left( {{A_1}} \right) + P\left( {A|{A_2}} \right) \cdot P\left( {{A_2}} \right)}}$
2.${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$, where $n$ is the total number of elements and $r$ is the number of elements to be selected

Complete step-by-step answer:
It is given that a pack of cards is facing downwards and when counted, it is found that one card is missing.
We have to find the probability that the missing card is red.
Now, we know that there are a total 52 cards.
Let $A$ be the event of drawing a red card when 1 card is missing i.e. the total cards are 51.
Let ${A_1}$ be the event of the missing card being red in colour.
Let ${A_2}$ be the event that the missing card is black in colour.
Now, we will use Bayes’ Theorem to find the required probability. Bayes’ Theorem is a mathematical formula used to determine the conditional probability of the given events.
Here, in the total pack of 52 cards; there are 26 red cards and 26 black cards respectively. Therefore, probability of the missing card being red will be
$P\left( {{A_1}} \right) = \dfrac{{{}^{26}{C_1}}}{{{}^{52}{C_1}}}$
Now using the formula ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$, we get
$\Rightarrow P\left( {{A_1}} \right) = \dfrac{{\dfrac{{26!}}{{1!\left( {25!} \right)}}}}{{\dfrac{{52!}}{{1!\left( {51!} \right)}}}}$
Computing the factorial and simplifying, we get
$\Rightarrow P\left( {{A_1}} \right) = \dfrac{{26}}{{52}} = \dfrac{1}{2}$
Similarly, we will find the probability of the missing card being black. Therefore,
$P\left( {{A_2}} \right) = \dfrac{{{}^{26}{C_1}}}{{{}^{52}{C_1}}}$
Now using the formula ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$, we get
$\Rightarrow P\left( {{A_2}} \right) = \dfrac{{\dfrac{{26!}}{{1!\left( {25!} \right)}}}}{{\dfrac{{52!}}{{1!\left( {51!} \right)}}}}$
Computing the factorial and simplifying, we get
$\Rightarrow P\left( {{A_2}} \right) = \dfrac{{26}}{{52}} = \dfrac{1}{2}$
Also, the probability of drawing a red card when the missing card is red will give us the total number of outcomes as 51 and the favorable outcomes as 25 (because we have to draw a red card only)
Therefore,
The probability of drawing a red card when the missing card is red $= P\left( {A|{A_1}} \right) = \dfrac{{25}}{{51}}$
Similarly, the probability of drawing a red card when the missing card is black will give us the total number of outcomes as 51 and the favorable outcomes as 26 (because we have to draw a red card and the missing card is black so it will have no effect)
Therefore,
The probability of drawing a red card when the missing card is black $= P\left( {A|{A_2}} \right) = \dfrac{{26}}{{51}}$
Now, substituting $P\left( {{A_1}} \right) = \dfrac{1}{2}$,$P\left( {{A_2}} \right) = \dfrac{1}{2}$ ,$P\left( {A|{A_1}} \right) = \dfrac{{25}}{{51}}$ and $P\left( {A|{A_2}} \right) = \dfrac{{26}}{{51}}$ in the formula $P\left( {{A_1}|A} \right) = \dfrac{{P\left( {A|{A_1}} \right) \cdot P\left( {{A_1}} \right)}}{{P\left( {A|{A_1}} \right) \cdot P\left( {{A_1}} \right) + P\left( {A|{A_2}} \right) \cdot P\left( {{A_2}} \right)}}$, we get,
$P\left( {{A_1}|A} \right) = \dfrac{{\dfrac{{25}}{{51}} \cdot \dfrac{1}{2}}}{{\dfrac{{25}}{{51}} \cdot \dfrac{1}{2} + \dfrac{{26}}{{51}} \cdot \dfrac{1}{2}}}$
Simplifying the expression, we get
$\Rightarrow P\left( {{A_1}|A} \right) = \dfrac{{25}}{{25 + 26}}$
Adding the terms in the denominator, we get
$\Rightarrow P\left( {{A_1}|A} \right) = \dfrac{{25}}{{51}}$
Therefore, the probability that the missing card is red is $\dfrac{{25}}{{51}}$.
Therefore, this is the required answer.

Note: The alternate way of solving this question is that, since, we know that the card drawn is red in colour and the total number of red cards is 26 in a pack of 52 cards. Hence, when a red card is drawn, then, 25 red cards are still left and the total number of cards also becomes 51. Hence, the probability that the missing card is red will have the total number of possible outcomes as 51 and the favorable outcomes as 25. Therefore, the required probability is $\dfrac{{25}}{{51}}$.
Hence, this is the required answer.