Question

There are $n$ different gift coupons, each of which can occupy $N(N>n)$ different envelopes, with the same probability $1/N$
$P_1$ : The probability that there will be one gift coupon in each of n definite envelopes out of $N$ given envelopes
$P_2$ : The probability that there will be one gift coupon in each of $n$ arbitrary envelopes out of $N$ given envelopes Consider the following statements
(i) $P_1=P_2$
(ii) $P_1=\frac{n!}{N^n}$
(iii) $P_2=\frac{N!}{N^n(N-n)!}$
(iv) $P_2=\frac{n!}{N^n(N-n)\text{ ! }}$
(v) $P_1=\frac{N\text{ ! }}{N^n}$
Now, which of the following is true

• A. Only (i)
• B. (ii) and (iii)
• C. (ii) and (iv)
• D. (iii) and (v)

Answer 1

Answer: (B)

From the given data $n(S)=N^n;n(A)=n!\Rightarrow P_1=\frac{n!}{N^n}$
$P_1=\frac{n!}{N^n}$ Since the $n$ different gift coupans can be placed in the $n$ definite (Out of $N$ ) envelope in ${}^n{P}_n=n$ ! ways
$P_2=\frac{N!}{(N-n)!N^n}$ As $n$ arbitrary envelopes out of $N$ given envelopes can be chosen in ${}^N{C}_n$ ways and the $n$ gift coupans can occupy these envelopes in $n$ ! ways. ]
$=\frac{N!}{N^n(N-n)!}$