Question

A bag contains 4 red, 3 blue and 2 white balls. Three balls are drawn at random. What is the probability that 2 are red and 1 is blue

Answer 1

ok so total number of balls = 4+3+2=9
we know that there are 9 balls in total
probability of 2 red balls = 2/4=1/2
probability of 1 blue ball = 1/3

it is the probability/the total number of balls in that particular colour

Answer 2

Answer:

Question:-

• A bag contains 4 red, 3 blue and 2 white balls. Three balls are drawn at random. What is the probability that 2 are red and 1 is blue?

$\bf \large\underbrace\orange{Answer:}$

Given:-

• A bag contains 4 red, 3 blue and 2 white balls and three balls are drawn at random.

To find :-

• Probability of 2 red and 1 blue.

Concept used :-

The use of Combinations along with probability.

$\bf \:\ ^{n}C_{r} = \dfrac{n!}{r!(n - r)!}$

$\bf \:Probability =\dfrac{Number \: of \: favourable \: outcomes}{Total \: number \: of \: outcomes}$

$\bf \large\underbrace\orange{Solution:}$

Bag have 4 red, 3 blue and 2 white balls.

⟹ Total number of balls = 4 + 3 + 2 = 9.

Step 1 :-

The number of ways in which three balls can be drawn from 9 balls is

$\bf \: ^{9}C_{3} = \dfrac{9!}{3!(9 - 3)!}$

$\bf\implies \: \bf \:\dfrac{9!}{3!×6!}$

$\bf\implies \:\dfrac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84 \: ways$

Step 2 :-

The number of ways in which 2 red balls and 1 blue ball can be selected from 4 red and 2 blue balls is

$\bf \:\ ^{4}C_{2} × \ ^{2}C_{1}$

$\bf\implies \ \: \dfrac{4!}{2!(4 - 2)!} \times \bf \:\dfrac{2!}{1!(2 -1) !}$

$\bf\implies \:\dfrac{4 \times 3}{2 \times 1} \times \dfrac{2 \times 1}{1}$

$\bf\implies \:12 \: ways$

Step :- 3

So required probability of getting 2 red and 1 blue ball is given by

$\bf \:Probability =\dfrac{Number \: of \: favourable \: outcomes}{Total \: number \: of \: outcomes}$

$\bf\implies \:\dfrac{12}{84} = \dfrac{1}{7}$

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