Question

A bag contains 6 white and 4 black balls .2 balls are drawn at random. Find the probability that they are of same colour.

Answer 1

the probablity of getting same ball is 5/6

Answer 2

==> The probability of getting the same colour is $P(A)=\frac{7}{15}$

Explanation:

Given:

1. A bag contains 6 white and 4 black balls

2. 2 balls are drawn at random

To find:

The probability that they are of the same colour.

Formula:

$^{n} C_{r} = \frac{n!}{r!(n-r)!}$

Probability  $P(A)=\frac{n(A)}{n(S)}$

PROBABILITY:

The probability of the event is defined as the ratio of the number of outcomes favourable to an event(A) to the total number of outcomes

Solution:

==> Total number outcomes, n(S)

==> drawing 2 balls out of 6 white balls and 4 black balls

==> 6 white balls + 4 black balls = 10 balls

==> n = 10

==> r = 2

==> $^{n} C_{r} = \frac{n!}{r!(n-r)!}$

==> $^{10} C_{2} = \frac{10!}{2!(10-2)!}$

==> $^{10} C_{2} = \frac{10!}{2!(8)!}$

==> $^{10} C_{2} = \frac{10\times9\times8\times7\times6\times5\times4\times3\times2\times1}{2!(8\times7\times6\times5\times4\times3\times2\times1)}$

==> $^{10} C_{2} = \frac{10\times9\times8\times7\times6\times5\times4\times3\times2\times1}{2!(8\times7\times6\times5\times4\times3\times2\times1)}$

==> $^{10} C_{2} = \frac{10\times9}{2\times1}$

==> $^{10} C_{2} = 5\times9$

==> ¹⁰C₂ = 45

==> n(S)=45

==> Event A = same color balls

==> 2 balls (from 6 white balls) or 2 balls(from 4 black balls)

==> 2 balls out of 6 white balls

==> n = 6

==> r = 2

==> $^{n} C_{r} = \frac{n!}{r!(n-r)!}$

==> $^{6} C_{2} = \frac{6!}{2!(6-2)!}$

==> $^{6} C_{2} = \frac{6!}{2!(4)!}$

==> $^{6} C_{2} = \frac{6\times5\times4\times3\times2\times1}{2!(4\times3\times2\times1)}$

==> $^{6} C_{2} = \frac{6\times5}{2\times1}$

==> $^{6} C_{2} = \frac{6\times5}{2\times1}$

==> ⁶C₂ = 3×5

==> ⁶C₂ = 15

==> 2 balls out of 4 black balls

==> $^{n} C_{r} = \frac{n!}{r!(n-r)!}$

==> $^{4} C_{2} = \frac{4!}{2!(4-2)!}$

==> $^{4} C_{2} = \frac{4!}{2!(2)!}$

==> $^{4} C_{2} = \frac{4\times3\times2\times1}{2!(2\times1)}$

==> $^{4} C_{2} = \frac{4\times3}{2\times1}$

==> ⁴C₂ = 2×3

==> ⁴C₂ = 6

==> 2 balls (from 6 white balls) or 2 balls(from 4 black balls)

==> ⁶C₂ + ⁴C₂

==> 15 + 6

==> n(A) = 21

==> $P(A)=\frac{n(A)}{n(S)}$

==> $P(A)=\frac{21}{45}$

==> $P(A)=\frac{7}{15}$

==> The probability of getting the same colour is $P(A)=\frac{7}{15}$