Question

A card is drawn at random from a deck of $52$ cards. Find the probability of drawing a heart or spade.

• A. $\frac{1}{4}$
  
• B. $\frac{1}{2}$
  
• C. $\frac{1}{13}$
  
• D. $\frac{1}{52}$
  

Answer 1

The correct option is B $\frac{1}{2}$


There are $13$ heart cards in a pack of $52$ cards.
Selection of $1$ heart card from $13$ heart cards can be done in ${}^{13}{C}_1$ ways.
Selection of a card from $52$ cards can be done in ${}^{52}{C}_1$ ways.

Let $\text{P(H)}$ denotes the probability of drawing a heart card.
Number of favorable outcomes $=$ Selecting a heart card from $13$ heart cards $={}^{13}{C}_1$
Total number of outcomes $=$ Selection of a card from $52$ cards $={}^{52}{C}_1$

$\text{P(H)}=\frac{\text{Selecting a heart card from 13 heart cards}}{\text{Selection of a card from 52 cards}}$
$\text{P(H)}=\frac{{}^{13}{C}_1}{{}^{52}{C}_1}$
$\text{P(H)}=\frac{13}{52}$
$\text{P(H)}=\frac{1}{4}$

There are $13$ spade cards in a pack of $52$ cards.
Selection of $1$ spade card from $13$ spade cards can be done in ${}^{13}{C}_1$ ways.
Selection of a card from $52$ cards can be done in ${}^{52}{C}_1$ ways.

Let $\text{P(S)}$ denotes the probability of drawing a spade card.
Number of favorable outcomes $=$ Selecting a spade card from $13$ spade cards $={}^{13}{C}_1$
Total number of outcomes $=$ Selection of a card from $52$ cards $={}^{52}{C}_1$

$\text{P(S)}=\frac{\text{Selecting a spade card from 13 spade cards}}{\text{Selection of a card from 52 cards}}$
$\text{P(S)}=\frac{{}^{13}{C}_1}{{}^{52}{C}_1}$
$\text{P(S)}=\frac{13}{52}$
$\text{P(S)}=\frac{1}{4}$

The events $\text{Drawing a heart card}$ and $\text{Drawing a spade card}$ are mutually exclusive events because they cannot occur at the same time.

Let $\text{P(H or S)}$ denotes the probability of drawing a heart or spade.

$\text{P(H or S)}=\text{P(H)}+\text{P(S)}$
$=\frac{1}{4}+\frac{1}{4}$
$\text{P(H or S)}=\frac{1}{2}$

The probability of drawing a heart or spade from a deck of $52$ cards is $\frac{1}{2}$.

Therefore, option (b.) is the correct answer.