Question

There are three clubs A, B, C in a town with 40, 50, 60 members respectively 10 people are members of all three clubs, 70 members belong to only one club. A member is randomly selected. Find the probability that he has membership in clubs.

Answer 1

Hint:
Firstly, find the number of people having memberships of exactly two clubs.
Then, find the total number of people having memberships.
Finally, to find the probability of a person having exactly two clubs use the formula $P = \dfrac{{Number\,of\,people\,having\,exactly\,two\,memberships}}{{Total\,number\,of\,people\,having\,memberships}}$ .

Complete step by step solution:
There are a total of 3 clubs in the city. Total number of members in three clubs is 40 + 50 + 60 = 150.
Also, the number of members that belong to one club only is 70. So, the number of remaining members is 150 – 70 = 80.
Now, as it is given, the number of people that belong to all the three clubs is 10. So, the number of people having all the three memberships is \[3\left( {10} \right)\] = 30.
So, the number of remaining members is 80 – 30 = 50.
Thus, the number of people having exactly two memberships is 50.
So, the number of people in each of the two clubs is $\dfrac{{50}}{2}$ =25.
Now, the equation to find the probability of a person having exactly two memberships is given as $P = \dfrac{{Number\,of\,people\,having\,exactly\,two\,memberships}}{{Total\,number\,of\,people\,having\,memberships}}$ .
$\therefore P = \dfrac{{25}}{{150}} = \dfrac{1}{6}$

Thus, the probability of a person having exactly two memberships is $\dfrac{1}{6}$.

Note:
Here, it is given that 10 people have the memberships of all three clubs, so we have to multiply the number with 3 to get the number of people having memberships and carry further. If we do not take the product of 10 and 3, and carry the question to solve further, we will not be able to get the required answer. Thus, take care to take the product of 10 and 3 and solve.