Question

A card is drawn from a pack of \[52\] cards. The card is drawn at random; find the probability that it is neither club nor queen?

Answer 1

Hint:
In this question, we have the total number of cards is\[52\]. And we know the total number of clubs and queens is $16$. So for neither club nor queen we will subtract from the total and then by using probability formula we will get the outcomes.

Formula used:
Suppose there is an event named as $A$ then the probability of an event will be
$P\left( A \right) = \dfrac{{{\text{Number of favourable outcome}}}}{{Total{\text{ number of favourable cases}}}}$
Or we can also write it as
$ \Rightarrow P\left( A \right) = \dfrac{{n\left( A \right)}}{{n\left( S \right)}}$
Here,
$P\left( A \right)$, will be the probability of an event $A$.
$n\left( A \right)$, will be the number of favorable outcomes.
$n\left( S \right)$, will be the total number of favorable cases.

Complete step by step solution:
So we have the cards from the pack and as we know the total number of cards is\[52\]in the pack. So the total number of favorable cases will be $n\left( S \right)$and can be written as
$ \Rightarrow n\left( S \right) = 52$
Now as we already know that the total number of club or queen cards in the pack will be
$ \Rightarrow 13 + 3 = 16$
So let us assume $E$will be the favorable outcomes of getting neither club nor queen
Therefore, $n\left( E \right) = 52 - 16$
And hence $n\left( E \right) = 36$
So by using the probability formula,
$ \Rightarrow P\left( E \right) = \dfrac{{n\left( E \right)}}{{n\left( S \right)}}$
Now on substituting the values, we get
$ \Rightarrow P\left( E \right) = \dfrac{{36}}{{52}}$
And on solving the above, we get
$ \Rightarrow P\left( E \right) = \dfrac{9}{{13}}$

Therefore, $\dfrac{9}{{13}}$ will be the probability that it is neither club nor queen.

Note:
For solving the probability question related to cards we have to be known with all the cards identities and one more thing that with practice, read and ask one will be good at probability. There is no other way to be best in this.