Question

Two cards are drawn successively without replacement from a well-shuffled deck of 52 cards . Find the probability distribution of the number of aces.

Answer 1

Hint: \[{}^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}\] . There will be three cases for the probability distribution There are 4 aces in a deck of 52 cards so the combination could be we get 0 aces, 1 ace or maximum of 2 aces.

Complete step-by-step answer:
The number of aces in a pack of cards are 4.
Let X be the discrete random variable denoting the number of aces when two cards are drawn without replacement.
Therefore X can take the value of 0,1 or 2
So let us take 3 cases
For case 1: \[X = 0\]
\[P(X = 0) = \dfrac{{{}^{48}{C_2} \times {}^4{C_0}}}{{{}^{52}{C_2}}}\]
And as we know that \[{}^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}\]
\[\therefore P(X = 0) = \dfrac{{1128 \times 1}}{{1326}} = \dfrac{{188}}{{221}}\]
Now for Case 2: The probability of 1 ace to be drawn i.e., \[X = 1\]
\[\begin{array}{l}
P(X = 1) = \dfrac{{{}^{48}{C_1} \times {}^4{C_1}}}{{{}^{52}{C_2}}}\\
 \Rightarrow P(X = 1) = \dfrac{{48 \times 4}}{{1326}} = \dfrac{{32}}{{221}}
\end{array}\]
For Case 3: Similarly the probability of 2 aces will be when \[X = 2\]
\[\begin{array}{l}
P(X = 2) = \dfrac{{{}^{48}{C_0} \times {}^4{C_2}}}{{{}^{52}{C_2}}}\\
 \Rightarrow P(X = 2) = \dfrac{{1 \times 6}}{{1326}} = \dfrac{1}{{221}}
\end{array}\]
Now let us try to draw the table of Probability distribution

X	P(X)
0	\[\dfrac{{188}}{{221}}\]
1	\[\dfrac{{32}}{{221}}\]
2	\[\dfrac{1}{{221}}\]

Note: Drawing the table for probability distribution is very much important also for \[{\bf{P}}\left( {{\bf{X}} = {\bf{0}}} \right)\] the combination i have used is we will pick any 2 cards from 48 because the rest 4 are aces and also we will choose 0 from the rest of the 4, that's why it was \[{}^{48}{C_2} \times {}^4{C_0}\] in the numerator which was ultimately divided by the total choices we have to take 2 cards from 52, i.e., \[{}^{52}{C_2}\] . This trick was followed for all the cases that's why the value of r in \[{}^n{C_r}\] was gradually growing as n remained the same.