Question

There are ten prizes, five $A$'s, three $B$'s and two $C$'s, placed in identical sealed envelopes for the top ten contestants in a mathematics contest. The prizes are awarded by allowing winners to select an envelope at random from those remaining. When the $8^{th}$ contestant goes to select the prize, the probability that the remaining three prizes are one $A$, one $B$ and one $C$, is

• A. $\frac{1}{4}$
• B. $\frac{1}{3}$
• C. $\frac{1}{12}$
• D. $\frac{1}{10}$

Answer 1

The correct option is A $\frac{1}{4}$

Given, In a competition there are $5A,3B,2C$ prizes.

Total no.of ways that $10$ participants can be alloted these prizes$=\frac{10!}{(5!\times 3!\times 21)}$

(The denominator part $5!\times 3!\times 2!$ has to do, since $5A,3B,2C$ repetition occuring)

Total no. of ways of selecting $7$ prizes such that when selecting 8th prize only $1A,1B,1C$ to left=$(\frac{7!}{(4!\times 2!\times 1!)})\times 3!$

$Intheabove(\frac{7!}{(4!\times 2!\times 1!)})\to$ No.of ways of selecting $4A,2B,1C$

$3!\to$indicates the no. of ways alloting remaining $1A,1B,1C$ prizes to the participant.

$\therefore Therequiredprobability=\frac{\frac{10!}{(5!\times 3!\times 21)}}{(\frac{7!}{(4!\times 2!\times 1!)})\times 3!}=\frac{(5\times 3\times 2)\times 3!}{(10\times 9\times 8)}=\frac{1}{4}$