Question

In a box, 4 coins of ten rupees, 2 coins of
five rupees, 2 coins of two rupees and 2
coins of one rupee are put. Now, three coins
are taken out randomly. What is the
probability that the amount drawn is 12
rupees?​

Answer 1

$\large\underline{\sf{Solution-}}$

➢ Given that a box contains

• 4 coins of ten rupees

• 2 coins of five rupees

• 2 coins of two rupees

• 2 coins of one rupee

Thus,

• Total number of coins = 4 + 2 + 2 + 2 = 10

So,

Total number of ways in which 3 coins can be taken out is

$\rm \: = \: \: ^{10}C_3$

$\rm \: = \: \: \dfrac{10!}{3! \: 7!}$

$\rm \: = \: \: \dfrac{10 \times 9 \times 8 \times 7!}{3 \times 2 \times 1 \: 7!}$

$\rm \: = \: \: 120 \: ways$

Now,

Amount of 12 rupees can be drawn in the following cases:

Case :- 1

➢ 1 coin of ten rupees and 2 coins of 1 rupee

The number of ways in which 1 coin of 10 rupees and 2 coins of 1 rupee can be drawn out of 4 coins and 2 coins is

$\rm \: = \: \: ^{4}C_1 \times \: ^{2}C_2$

$\rm \: = \: \: \dfrac{4!}{1! \: 3!} \times \dfrac{2!}{2! \: 0!}$

$\red{\bigg \{ \because \:^{n}C_r \: = \: \dfrac{n!}{r! \: (n - r)!} \bigg \}}$

$\rm \: = \: \: \dfrac{4 \times 3!}{ \: 3!}$

$\rm \: = \: \: 4 \: ways$

Case :- 2

➢ 2 coins of five rupees and 1 coin of 2 rupees

The number of ways in which 2 coin of 5 rupees and 1 coins of 2 rupee can be drawn out of 2 coins and 2 coins is

$\rm \: = \: \: ^{2}C_2 \times \: ^{2}C_1$

$\rm \: = \: \: \dfrac{2!}{2! \: 0!} \times \dfrac{2!}{1! \: 1!}$

$\rm \: = \: \: \dfrac{2\times 1}{1}$

$\rm \: = \: \: 2$

Hence,

➢ Total number of ways in which 12 rupees can be drawn is = 4 + 2 = 6.

We know,

$\sf \:Probability \: of \: an \: event =\dfrac{Number \: of \: favourable \: outcomes}{Total \: number \: of \: outcomes \: in \: sample \: space}$

So,

$\rm :\longmapsto\: \:P(getting \: amount \: of \: 12 \: rupees) = \dfrac{6}{120}$

$\rm :\longmapsto\: \:P(getting \: amount \: of \: 12 \: rupees) = \dfrac{1}{20}$