Question

Two cards are drawn simultaneously from a well-shuffled pack of 52 cards. Find the probability distribution of the number of jacks.

Answer 1

Here, we will take into consideration cases such that when two cards are drawn then either both of them can be a jack or either of them can be a jack or none of them can be a jack. We will consider the values 0, 1 and 2 for this probability distribution. Then we will find the probabilities of getting a jack in each case and then put them in the table to get the required probability distribution of the number of jacks.

Formula Used:
Probability, $P =$ Favorable outcomes $\div$ Total number of outcomes

Complete Step by step Solution:
It is given that we have a well-shuffled pack of 52 cards.
Now, two cards are drawn without replacement.
Let $X$ denote the number of jacks in the sample space of 2 cards which are drawn from the well-shuffled pack of 52 cards.
Hence, either we can have 1 jack, 2 jacks or no jack at all.
Therefore,$X$ can take the values 0, 1 and 2.
Now,
Case 1: When none of the two cards contain a jack.
Hence, $X = 0$
Now, we know that the total number of jacks in a pack of 52 cards is 4.
Hence, the remaining cards or the favorable outcomes $= 52 - 4 = 48$
And the total number of outcomes $= 52$
Substituting these values in the formula Probability $P =$ Favorable outcomes $\div$ Total number of outcomes, we get
Now, when the first card is drawn, which is not a jack,
Probability, $P = \dfrac{{48}}{{52}}$
Since, one card has been drawn, hence, the probability of drawing second card will become:
$P = \dfrac{{47}}{{51}}$
Hence, total probability of the cards not being a jack,
$P\left( {X = 0} \right) = \dfrac{{48}}{{52}} \times \dfrac{{47}}{{51}} = \dfrac{{4 \times 47}}{{17 \times 13}} = \dfrac{{188}}{{221}}$
Case 2: When one of the two cards contains a jack.
Hence, $X = 1$
The total number of jacks in a pack of 52 cards $= 4$
Hence, the remaining cards or the favorable outcomes $= 52 - 4 = 48$
And the total number of outcomes $= 52$
Substituting these values in the formula Probability $P =$ Favorable outcomes $\div$ Total number of outcomes, we get
Now, probability of first card being a jack $= \dfrac{4}{{52}}$
And the probability of the next card not being a jack when 1 card is drawn$= \dfrac{{48}}{{51}}$
But, this could be in reverse as well, i.e. the first card could be any of the remaining cards, hence giving us the probability as $\dfrac{{48}}{{52}}$
And the second card could be a jack, hence, probability will be $\dfrac{4}{{51}}$
Hence, total probability of one of cards being a jack,
$P\left( {X = 1} \right) = \dfrac{4}{{52}} \times \dfrac{{48}}{{51}} + \dfrac{{48}}{{52}} \times \dfrac{4}{{51}}$
$\Rightarrow P\left( {X = 1} \right) = 2\left( {\dfrac{1}{{13}} \times \dfrac{{16}}{{17}}} \right) = 2\left( {\dfrac{{16}}{{221}}} \right) = \dfrac{{32}}{{221}}$
Case 3: When both the cards contain a jack.
Hence, $X = 2$
Now, we know that total number of jacks in a pack of 52 cards $= 4$
Hence, the remaining cards or the favorable outcomes $= 52 - 4 = 48$
And the total number of outcomes $= 52$
The formula of Probability, $P =$ Favorable outcomes $\div$ Total number of outcomes
Now, probability of first card being a jack $= \dfrac{4}{{52}}$
And the probability of the next card also being a jack when 1 jack is already drawn $= \dfrac{3}{{51}}$
Hence, total probability of both the cards being a jack,
$P\left( {X = 2} \right) = \dfrac{4}{{52}} \times \dfrac{3}{{51}}$
$\Rightarrow P\left( {X = 2} \right) = \dfrac{1}{{13}} \times \dfrac{1}{{17}} = \dfrac{1}{{221}}$
Therefore, the required probability distribution of $X$ or the number of jacks is given by:

$X$	0	1	2
$P\left( X \right)$	$\dfrac{{188}}{{221}}$	$\dfrac{{32}}{{221}}$	$\dfrac{1}{{221}}$

Hence, this is the required probability distribution of the number of jacks.

Note:
Probability tells us the extent to which an event is likely to occur, i.e. the possibility of the occurrence of an event. Hence, it is measured by dividing the favorable outcomes by the total number of outcomes. Probability of an event cannot be less than 0 or it cannot be greater than 1. This means the probability of an event lies between 0 to 1. When the probability is 1 then the event is called a sure event, whereas when the probability is 0 the event will not occur.