Question

A card is drawn from a well-shuffled deck of 52 cards. How do you find the P(drawing an ace or a 9) ?

Answer 1

Hint: To solve this question we will use the basic formula of probability of an event to occur which is given by $P\left( A \right)=\dfrac{n\left( E \right)}{n\left( S \right)}$
Where, A is an event which we want to occur \[n\left( E \right)=\] Number of favorable outcomes and $n\left( S \right)=$ number of total possible outcomes
We will calculate the individual probability of events drawing an ace and a 9. Then we will combine both to get the desired answer.

Complete step by step solution:
We have been given that a card is drawn from a well-shuffled deck of 52 cards.
We have to find the probability of drawing an ace or a 9.
Now, we know that the probability of an event A to occur is given as $P\left( A \right)=\dfrac{n\left( E \right)}{n\left( S \right)}$
Where, A is an event which we want to occur
\[n\left( E \right)=\] Number of favorable outcomes and $n\left( S \right)=$ number of total possible outcomes
Now, we have a total 52 cards so the total number of possible outcomes will be the same in both the cases.
So we get $n\left( S \right)=52$
Now, let us assume that ${{E}_{1}}$ is the event of drawing an ace.
As we know that there are 4 aces in the deck of 52 cards so we get \[n\left( {{E}_{1}} \right)=4\]
So the probability of drawing an ace will be
 $\Rightarrow \dfrac{4}{52}$
Now, let us assume that ${{E}_{2}}$ is the event of drawing a 9.
As we know that there are 4 9’s in the deck of 52 cards so we get \[n\left( {{E}_{2}} \right)=4\]
So the probability of drawing a 9 will be
 $\Rightarrow \dfrac{4}{52}$
Now, we know that both the events are mutually exclusive and we have to find the probability of getting one event OR another so we need to use the OR rule which is given as
\[P\left( {{E}_{1}}or{{E}_{2}} \right)=P\left( {{E}_{1}} \right)+P\left( {{E}_{2}} \right)\]
Now, using the above rule and substituting the values we will get
$\begin{align}
  & \Rightarrow P\left( \text{drawing an ace or 9} \right)=\dfrac{4}{52}+\dfrac{4}{52} \\
 & \Rightarrow P\left( \text{drawing an ace or 9} \right)=\dfrac{4+4}{52} \\
 & \Rightarrow P\left( \text{drawing an ace or 9} \right)=\dfrac{8}{52} \\
 & \Rightarrow P\left( \text{drawing an ace or 9} \right)=\dfrac{2}{13} \\
\end{align}$

Note: The point to be noted is that when we calculate the probability of occurrence of any event the value should lie between 0-1. The value of probability cannot exceed one. Probability value higher than one means probability greater than $100\%$ and it is not possible.