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., $X = 1$

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.