Question

Coupons numbered $1,2,.....,25$ are mixed up and one coupon is drawn at random. What is the probability that the coupon has a number which is multiple of $2$ or $3$?
A.$\dfrac{{16}}{{25}}$
B.$\dfrac{1}{5}$
C.$\dfrac{3}{5}$
D.$\dfrac{{12}}{{25}}$

Answer 1

Hint : As given in the question, the coupons are numbered from $1$ to $25$. Thus we can say that the total number of coupons is $25$. Now these $25$ coupons are mixed and one coupon among these $25$ coupons is drawn randomly. Now we need to find the probability that the coupon number drawn is a multiple of $2$ or $3$. As we know there are many common multiples of $2$ and $3$ from $1$ to $25$. So, here we will apply the $P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right)$ formula.
Formula:
This is a joint probability question and formula to calculate joint probability is:
$P\left( {A{\text{ }}or{\text{ }}B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A{\text{ }}and{\text{ }}B} \right)$
Equivalently, $P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right)$

Complete step-by-step answer :
Sample space (set of all possible outcomes) = $25$
Let us write the multiples:
Multiples of $2$ from $1$ to $25$ are $2,4,6,8,10,12,14,16,18,20,22,24$
Multiples of $3$ from $1$ to $25$ are $3,6,9,12,15,18,21,24$
Common multiples between $2$ and $3$ are $6,12,18,24$
Now, we will find the probability using the $P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right)$ formula.
Here, $P\left( A \right)$ = Total number of multiples of $2$ / Sample space.
There are total $12$ multiples of $2$ from $1$ to $25$
$\Rightarrow P\left( A \right) = \dfrac{{12}}{{25}}$
Here, $P\left( B \right)$ = Total number of multiples of $3$ / Sample space.
There are total $8$ multiples of $3$ from $1$ to $25$
$\Rightarrow P\left( B \right) = \dfrac{8}{{25}}$
Here, $P\left( {A \cap B} \right)$ = Common favorable events between $A$ and $B$ / Sample space.
As we know there are four common events between $A$ and $B$, $6,12,18,24$
$\Rightarrow P\left( {A \cap B} \right) = \dfrac{4}{{25}}$
$P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right)$
Put the values of $P\left( A \right),P\left( B \right),P\left( {A \cap B} \right)$ in the above written formula.
$\Rightarrow P\left( {A \cup B} \right) = \dfrac{{12}}{{25}} + \dfrac{8}{{25}} - \dfrac{4}{{25}}$
Take LCM of $20$
$\Rightarrow P\left( {A \cup B} \right) = \dfrac{{12 + 8 - 4}}{{25}}$
$\therefore P\left( {A \cup B} \right) = \dfrac{{16}}{{25}}$
Thus the probability that the ticket drawn has a multiple of $2$ or $3$ = $\dfrac{{16}}{{25}}$
Therefore, option (A) is the correct answer.
So, the correct answer is “Option B”.

Note : This question can also be solved by set theory. Let $S$ be the total number of tickets, then
S = \left\\{ {1,2,3,4,.....,25} \right\\}
Let E be the event of getting a multiple of $2$ or $3$. Thus we get $E$ as,
E = \left\\{ {2,4,6,8,10,12,14,16,18,20,22,24,3,9,15,21} \right\\}
(A set cannot have duplicate/repeated elements that is why we will not write $6,12,18,24$again)
Now required probability = $P\left( E \right)$
$P\left( E \right) = \dfrac{{n\left( E \right)}}{{n\left( S \right)}} = \dfrac{{16}}{{25}}$