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 in an ace, is
A). $\dfrac{9}{{20}}$
B). $\dfrac{3}{{16}}$
C). $\dfrac{1}{6}$
D). $\dfrac{1}{9}$

Answer 1

Start by finding out the total number of cards available for selection, look for all the possible cases through which we can get at least one Ace and use a combination formula for selection. Apply the formula for probability and find the required value by adding the cases.

Complete step by step solution:
Given,
No. of Aces = 4
No. of Kings= 4
No. of Queens = 4
No. of Jacks= 4
So, the total number of cards we have = 16
Now, we need to select 2 cards out of these, which can be done in ${}^{16}{C_2}$ ways
Now, we need to find the probability of at least an ace.
And we know probability of an event = $\dfrac{{{\text{No}}{\text{. of favourable outcome}}}}{{{\text{Total no}}{\text{. of sample}}}} = \dfrac{{n(E)}}{{n(S)}}$
So , we can select at least one ace in two ways
Case1:- 1 Ace + 1 other card
1 Ace can be selected from 4 Aces in ${}^4{C_1}$ ways
1 other card from leftover 12 cards can be selected in ${}^{12}{C_1}$ways
So , selecting 1 Ace and 1 other card can be done in ${}^4{C_1} \times {}^{12}{C_1}$ways
Therefore, Probability of 1 Ace and 1 other card = $\dfrac{{{}^4{C_1} \times {}^{12}{C_1}}}{{{}^{16}{C_2}}}$
Case2:- both the cards are Ace
2 Aces can be selected from 4 aces in ${}^4{C_2}$ways
Probability of 2 Ace = $\dfrac{{{}^4{C_2}}}{{{}^{16}{C_2}}}$
Adding both the cases , we’ll get the probability of at least 2 aces
Therefore, the probability of at least 2 aces = $\dfrac{{{}^4{C_1} \times {}^{12}{C_1}}}{{{}^{16}{C_2}}}$+$\dfrac{{{}^4{C_2}}}{{{}^{16}{C_2}}}$
By using the formula ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ expanding all the terms.
$= \dfrac{{4 \times 12 \times 2! \times 14!}}{{16 \times 15 \times 14!}} + \dfrac{{4 \times 3 \times 2! \times 14!}}{{2! \times 16 \times 15 \times 14!}} \\\ = \dfrac{{4 \times 12 \times 2}}{{16 \times 15}} + \dfrac{{4 \times 3}}{{16 \times 15}} \\\ = \dfrac{2}{5} + \dfrac{1}{{20}} \\\ = \dfrac{9}{{20}}$
Therefore the probability of at least an Ace is $\dfrac{9}{{20}}$. So, option A is the correct answer.

Note: One must be thorough with all the formulas and concepts used in probability. Attention must be given while making the selection as it involves the use of Combination theory and relevant formulas. Product and addition rules in combination must be well understood.