Question

A pack of cards contains 4 aces, 4 kings, 4 queens, and 4 jacks. Two cards are drawn at random. The probability that at least one of these is an ace is
A.$\dfrac{9}{{20}}$
B.$\dfrac{3}{{16}}$
C.$\dfrac{1}{6}$
D.$\dfrac{1}{9}$

Answer 1

We will begin by finding the total number of cards in the pack. Then, we will consider two possibilities of drawing two cards in which at least one is an ace. Next, we will use the formula of combinations to find the probability in each case. Finally, we will find the total probability by adding the probabilities obtained in the two cases.

Formula used:
We will use the following formulas:
If A is an event, then $P(A) =$ no. of favourable outcomes $\div$ total no. of outcomes
$C(n,r) = {}^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}$
$n! = n \times (n - 1)! = n \times (n - 1) \times (n - 2)!$

Complete step-by-step answer:
It is given that the pack of cards contains 4 aces, 4 kings, 4 queens, and 4 jacks. So, there are 16 cards in total. Two cards are drawn and we must find the probability that at least one of these is an ace. We will consider two cases.
Case I: One of the cards is an ace and the other card can be either a king, queen or a jack.
Out of 4 aces, we are drawing one ace and out of the remaining 12 cards, we are drawing one card. Now, one ace can be drawn in ${}^4{C_1}$ ways.
The other card can be drawn in ${}^{12}{C_1}$ ways.
So, the possibility of drawing one ace and one other card is ${}^4{C_1} \times {}^{12}{C_1}$.
Now, drawing 2 cards out of 16 given cards can be done in ${}^{16}{C_2}$ ways. Hence,
$P$ (drawing one ace and one other card) $= \dfrac{{{}^4{C_1} \times {}^{12}{C_1}}}{{{}^{16}{C_2}}}$
Now, using $C(n,r) = {}^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}$, we have
$\Rightarrow$ $P$ (drawing one ace and one other card) $= \dfrac{{\dfrac{{4!}}{{1!(4 - 1)!}} \times \dfrac{{12!}}{{1!\left( {12 - 1} \right)!}}}}{{\dfrac{{16!}}{{2!(16 - 2)!}}}}$
Simplifying the expression, we get
$\Rightarrow$ $P$ (drawing one ace and one other card) $= \dfrac{{\dfrac{{4 \times 3!}}{{1 \times 3!}} \times 12}}{{\dfrac{{16 \times 15 \times 14!}}{{2! \times 14!}}}}$
Computing the factorial, we get
$\therefore$ $P$ (drawing one ace and one other card) $= \dfrac{{4 \times 12}}{{120}} = \dfrac{2}{5}$

Case II: Out of 4 aces, we are drawing 2 aces.
This can be done in ${}^4{C_2}$ ways.
Drawing 2 cards out of 16 given cards can be done in ${}^{16}{C_2}$ ways. Hence,
$P$ (drawing two aces) $= \dfrac{{{}^4{C_2}}}{{{}^{16}{C_2}}} = \dfrac{{\dfrac{{4!}}{{2!(4 - 2)!}}}}{{120}}$
Simplifying the expression, we get
$\Rightarrow$ $P$ (drawing two aces) $= \dfrac{{\dfrac{{4!}}{{2!2!}}}}{{120}}$
Computing the factorial, we get
$\Rightarrow$ $P$ (drawing two aces) $= \dfrac{{\dfrac{{4 \times 3 \times 2!}}{{2!2!}}}}{{120}}$
Simplifying the expression further , we get
$\Rightarrow$ $P$ (drawing two aces) $= \dfrac{1}{{20}}$
Now we will find the probability of drawing at least one ace by combining the above cases. Therefore, we have
$P$ (drawing at least one ace) $= \dfrac{2}{5} + \dfrac{1}{{20}}$
Taking LCM, we get
$\Rightarrow$ $P$ (drawing at least one ace) $= \dfrac{{8 + 1}}{{10}}$
Adding the terms in the numerator, we get
$\Rightarrow$ $P$ (drawing at least one ace) $= \dfrac{9}{{20}}$
Hence, the correct option is A.

Note: We might get confused between combinations and Permutations. In permutations, the order of elements is to be taken care of, whereas in combinations, the order of elements does not matter. Here, the order in which we draw the cards is not a considerable factor, and this is the reason we have used combinations.