Question

A card is drawn from a deck of 52 cards. Find the probability of getting an ace or a spade card.
(a) $\dfrac{4}{13}$
(b) \[\dfrac{4}{26}\]
(c) \[\dfrac{2}{13}\]
(d) \[\dfrac{2}{26}\]

Answer 1

Hint: In the problem, we are asked to find the probability of getting an ace or a spade is asked. Therefore, we need to find the union of both events. The formula for finding the union of two events is $P\left( A\cup B \right)=P\left( A \right)+P\left( B \right)-P\left( A\cap B \right)$

Complete step by step solution:
Let the event of getting an ace card be \[A\] and the event of getting a spade be \[B\].
The probability of occurring of event \[A\] is \[P\left( A \right)\]and similarly, the probability of occurring of event\[B\]is \[P\left( B \right)\].
There are 4 aces in a deck of cards (ace of heart, spade, diamond, club).
Therefore, the probability of the event \[A\] is \[P\left( A \right)=\dfrac{\text{Number of favourable outcome}}{\text{Total number of favourable outcomes}}\]
Therefore, \[P\left( A \right)=\dfrac{4}{52}.............\left( i \right)\]
There are 13 cards of each type in the deck of cards (heart, spade, diamond, club).
Therefore, the probability of the event \[B\] is \[P\left( B \right)=\dfrac{\text{Number of favourable outcome}}{\text{Total number of favourable outcomes}}\]
Therefore, \[P\left( B \right)=\dfrac{13}{52}.............\left( ii \right)\]
There is one card which comes under event \[A\] as well as event $B$. So we need to make sure that we don’t count repeatedly.
Therefore, the intersection of event \[A\] and event \[B\] is \[A\cap B\] = Number of a favourable outcome.
Therefore, \[A\cap B=1\].
The probability of the event \[A\cap B\] is \[P\left( A\cap B \right)=\dfrac{\text{Number of favourable outcome}}{\text{Total number of favourable outcomes}}\].
Therefore, $P\left( A\cap B \right)=\dfrac{1}{52}....................\left( iv \right)$.
By using the formula, $P\left( A\cup B \right)=P\left( A \right)+P\left( B \right)-P\left( A\cap B \right)$
From $\left( i \right),\left( ii \right),\left( iii \right),\left( iv \right)$and substituting the values in the above equation, we get,
$P\left( A\cup B \right)=\dfrac{4}{52}+\dfrac{13}{52}-\dfrac{1}{52}=\dfrac{16}{52}$
Simplifying the equation,
$P\left( A\cup B \right)=\dfrac{16}{52}=\dfrac{4}{13}$
Hence, $P\left( A\cup B \right)=\dfrac{4}{13}$is the required solution and option (a) is correct.

Note: It is important to subtract the intersection term as it will be counted twice. Whenever the probability of finding an event \[A\] or $B$ is asked we need to find the union and whenever the probability of finding an event \[A\] and $B$ is asked we need to find the intersection. The main key in this problem is to remember the formula and be aware of how the deck of cards works.