Question

A box contains 5 green, 4 yellow and 3 white marbles. Three marbles are drawn at random
What is the probability that they are not of the same colour?
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Answer 1

Answer-

Probability that they are not of the same colour is

$\frac{1}{22}$

Hope it helps

Answer 2

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

➢Given that, box contains

• Number of green marbles= 5

• Number of yellow marbles = 4

• Number of white marbles = 3

Thus,

• Total number of marbles in box = 5 + 4 + 3 = 12.

Now,

➢Number of ways in which 3 marbles can be drawn randomly from 12 marbles is

$\rm \: = \: \: ^{12}C_{3}$

$\rm \: = \: \: \dfrac{12!}{3! \: (12 - 3)!}$

$\rm \: = \: \: \dfrac{12!}{3! \:9!}$

$\rm \: = \: \: \dfrac{12 \times 11 \times 10 \times 9!}{3 \times 2 \times 1\times \:9!}$

$\rm \: = \: \: 660$

So, Number of ways in which 3 marbles are drawn from 12 marbles = 660 ways.

Now,

➢Number of ways to draw 1 green marble from 5 green marbles is

$\rm \: = \: \: ^{5}C_{1}$

$\rm \: = \: \: \dfrac{5!}{1! \: (5 - 1)!}$

$\rm \: = \: \: \dfrac{5!}{\: 4!}$

$\rm \: = \: \: \dfrac{5 \times 4!}{\: 4!}$

$\rm \: = \: \: 5$

Now,

➢Number of ways to draw 1 yellow marble from 4 yellow marbles is

$\rm \: = \: \: ^{4}C_{1}$

$\rm \: = \: \: \dfrac{4!}{1! \: (4 - 1)!}$

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

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

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

Again,

➢Number of ways to draw 1 white marble from 3 white marbles is

$\rm \: = \: \: ^{3}C_{1}$

$\rm \: = \: \: \dfrac{3!}{1! \: (3 - 1)!}$

$\rm \: = \: \: \dfrac{3!}{\: 2!}$

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

$\rm \: = \: \: 3$

Hence,

➢ Number of ways in which 3 marbles can be drawn, not of same color is = 5 × 4 × 3 = 60.

Now,

We know,

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

➢So, Probability of getting 3 marbles of different color is

$\rm \: = \: \: \dfrac{60}{660}$

$\rm \: = \: \: \dfrac{1}{11}$