Question

A card is drawn at random from a well-shuffled pack of $52$ cards. Find the probability of getting a red card or a diamond or a jack ?

• A. $\frac{10}{13}$
• B. $\frac{15}{26}$
• C. $\frac{41}{52}$
• D. $\frac{7}{13}$

Answer 1

The correct option is D

$\frac{7}{13}$

Explanation for the correct option:

Finding $n\left(S\right),n\left(A\right),n\left(B\right),\mathrm{and}n\left(C\right)$

Let $S$ be the number of ways $2\mathrm{cards}$ can be drawn from $1$ pack of $52\mathrm{cards}$.

$n\left(S\right)={}^{52}C_1$

$=\frac{52!}{1!51!}$ $\left[{}^{n}C_r=\frac{n!}{r!\left(n-r\right)!}\right]$

$=\frac{52\times 51!}{51!}=52$

Let $A$ be the event that the card drawn is a $\mathrm{red}\mathrm{cards}$. Total $26\mathrm{red}\mathrm{cards}$ in a pack of cards,

$\begin{array}{rcl}n\left(A\right)& =& {}^{26}C_1\\ & =& \frac{26!}{1!25!}\\ & =& \frac{26\times 25!}{25!}\\ & =& 26\end{array}$

Let $B$ be the event, the card drawn is a diamond. Total $13$ diamond cards in a pack of cards,

$\begin{array}{rcl}n\left(B\right)& =& {}^{13}C_1\\ & =& \frac{13!}{1!12!}\\ & =& \frac{13\times 12!}{12!}\\ & =& 13\end{array}$

Let $C$ be the event, the card drawn is a jack. Total $4$ jacks in a pack of cards,

$\begin{array}{rcl}n\left(C\right)& =& {}^{4}C_1\\ & =& \frac{4!}{1!3!}\\ & =& \frac{4\times 3!}{3!}\\ & =& 4\end{array}$

Finding $n\left(A\cap B\right),n\left(B\cap C\right),n\left(A\cap C\right),\mathrm{and}n\left(A\cap B\cap C\right)$.

The chosen card can be a red as well as a diamond,

$\Rightarrow n\left(A\cap B\right)={}^{13}C_1=13$

The chosen card can be a diamond as well as a jack,

$\Rightarrow n\left(B\cap C\right)={}^{1}C_1=1$

The chosen card can be a red card as well as a jack,

$\Rightarrow n\left(A\cap C\right)={}^{2}C_1=2$

The chosen card can be a red card and a diamond and a jack,

$\Rightarrow n\left(A\cap B\cap C\right)={}^{1}C_1=1$

Finding the required probability.

The required probability $=P\left(A\cup B\cup C\right)$

$\begin{array}{rcl}P\left(A\cup B\cup C\right)& =& P\left(A\right)+P\left(B\right)+P\left(C\right)-P\left(A\cap B\right)-P\left(B\cap C\right)-P\left(A\cap C\right)-P\left(A\cap B\cap C\right)\\ & =& \frac{n\left(A\right)}{n\left(S\right)}+\frac{n\left(B\right)}{n\left(S\right)}+\frac{n\left(C\right)}{n\left(S\right)}-\frac{n\left(A\cap B\right)}{n\left(S\right)}-\frac{n\left(B\cap C\right)}{n\left(S\right)}-\frac{n\left(A\cap C\right)}{n\left(S\right)}+\frac{n\left(A\cap B\cap C\right)}{n\left(S\right)}\\ & =& \frac{26}{52}+\frac{13}{52}+\frac{4}{52}-\frac{13}{52}-\frac{1}{52}-\frac{2}{52}+\frac{1}{52}\\ & =& \frac{28}{52}=\frac{7}{13}\end{array}$

Hence, the correct answer is option D.