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 8th contestant goes to select the prize, the probability that the remaining three prizes are one A, one B and one C, is
A) $\dfrac{1}{4}$
B) $\dfrac{1}{3}$
C) $\dfrac{1}{{12}}$
D) $\dfrac{1}{{10}}$

Answer 1

Here we will firstly find the number of ways in which these prizes can be distributed among the 10 contestants. Then we will find the value of the number of ways of selecting the first 7 prizes selected such that the last three prizes left will be 1 A and 1 B and 1 C. Then we will find the ratio of the number of ways we calculated earlier to get the required probability.

Complete Step by Step Solution:
It is given that in a contest there are five A's, three B's and two C's.
Firstly we will find the probability of the number of ways in which these prizes can be distributed among the 10 contestants. Therefore, we get
Number of ways in which these prizes can be distributed among the 10 contestants is equal to $\dfrac{{10!}}{{5! \times 3! \times 2!}}$…………………………$\left( 1 \right)$
Now we will find the number of ways in which the first 7 prizes are selected such that the last three prizes left will be 1 A and 1 B and 1 C. Therefore, we get
Number of ways of selecting the first 7 prizes is selected such that the last three prizes left will be 1 A and 1 B and 1 C is $\dfrac{{7!}}{{4! \times 2! \times 1!}} \times 3!$…………………………$\left( 2 \right)$
Now by simply finding the ratio of the equation $\left( 1 \right)$ to the equation $\left( 2 \right)$ we will get the required probability. Therefore, we get
The required probability $= \dfrac{{\dfrac{{10!}}{{5! \times 3! \times 2!}}}}{{\dfrac{{7!}}{{4! \times 2! \times 1!}} \times 3!}}$
Now by solving the above equation we get
$\Rightarrow$ The required probability $= \dfrac{{\left( {5 \times 3 \times 2} \right) \times 3!}}{{\left( {10 \times 9 \times 8} \right)}} = \dfrac{1}{4}$
Hence when the ${8^{{\rm{th}}}}$ contestant goes to select the prize, the probability that the remaining three prizes are one A, one B, and one C is $\dfrac{1}{4}$.

So, option A is the correct option.

Note:
Here we should note that for finding the number of ways for the particular case by the use of the permutation not combination. The permutation is defined as the different ways in which a collection of items can be arranged. Combinations may be defined as the various ways in which objects from a set may be selected. Probability is the branch of mathematics which gives the possibility of the event occurrence and is equal to the ratio of the number of favorable outcomes to the total number of outcomes. Probability generally lies between the values of 0 to 1.