If a stamp vendor sells tickets of Rs 1 each and there are 3 persons having Rs 1 coin and 3 having Rs 2 coin standing in a row. Then the probability that stamp vendor do not run out of change if he does not have any money to start with is (assume each person will buy 1 ticket)(a) $\dfrac{1}{4}$ (b) $\dfrac{1}{2}$ (c) $\dfrac{3}{4}$ (d) None of these

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Hint: It is given that there are two types of buyers, buyers with Rs 1 coin and buyers with Rs 2 coins. So, while finding the arrangements, two types will be considered and not six people. First of all, we will find the sample set, in which the buyers can be arranged in any way. Then, we will find the number of ways the buyers can be arranged so that the stamp vendor always has Rs 1 coin to return to the person who has Rs 2 coin. In this way he will never be out of change. The probability can be found as the quotient of the number of arrangements of people in which he always has changed and the total number of arrangements.

It is given that there are 6 people in a row. Three have Rs 1 coin and three have Rs 2 coin.
Let S be the event of the buyers standing in any order.
There are 6 people in a row, 3 of one type and 3 of another.
Therefore, the number of arrangements will be $\dfrac{6!}{3!3!}$ = 20.
Let the number of elements in event set S be n(S). Thus, n(S) = 20.
Now, we need to find the arrangements in which the stamp vendor never runs out of change. This means, for every buyer having Rs 2 coin, there must be a buyer with Rs 1 coin who has already bought the stamp.
Let us denote buyers with Rs 1 coin as a and buyers with Rs 2 coin as b.
The following are the arrangements from left to right such that every b has a counterpart a who has already bought a ticket.
(i) a a a b b b
(ii) a b a b a b
(iii) a a b b a b
(iv) a b a a b b
(v) a a b a b b
So, if A is the event such that the stamp vendor doesn’t run out of change, then the number of such arrangements is n(A) = 5.
The probability can be found as the quotient of the number of arrangements of people in which he always has changed and the total number of arrangements.
$\Rightarrow$ P(A) = $\dfrac{n\left( \text{A} \right)}{n\left( \text{S} \right)}$
$\Rightarrow$ P(A) = $\dfrac{5}{20}$
$\Rightarrow$ P(A) = $\dfrac{1}{4}$
So, the correct answer is “Option A”.

Note: If we have n things which we have to arrange in n places, but r of one type, q of another type and rest all are different, then the number of arrangements will be $\dfrac{n!}{r!q!}$ . This is a concept from permutations and combinations. The chapter of permutation and combinations is a prerequisite for probability.