Question

A bag contains 3 red, 2 green and 4 blue balls. if 2 balls are randomly drawn, find the probability that both are of same colour. . 2/9 5/18 1/36 none of these

Answer 1

none of these
the probability of the event of getting both balls of the same color  = 10/9

Answer 2

Concept:

Probability is solely how likely something is to happen. The probability formula is defined because the possibility of a happening to happen is up to the ratio of the quantity of favourable outcomes and also the total number of outcomes.

Given:

we are providing that a bag has $3$ red, $2$ green and $4$ blue balls.

Find:

we have to search out the probability that the random balls which are drawn both are of same colour.

Solution:

Firstly, we'll find the total number of balls, we get

Total balls $=3+2+4=9$

Now, we'll find the quantity of favorable cases that the balls drawn from same colour is $^3C_{2}+^2C_{2}+^4C_{2}$.

To simplify the favourable cases we'll apply the formula $^nC_{r}=\frac{n!}{(n-r)!r!}$, we get

$\begin{aligned}\text{favourable cases}&=\frac{3!}{(3-2)!2!}+\frac{2!}{(2-2)!2!}+\frac{4!}{(4-2)!2!}\\ &=\frac{3!}{1!2!}+\frac{2!}{0\times 2!}+\frac{4!}{2!2!}\\ &=3+1+6\\ &=9\end$

Further, we are going to find the entire number of cases, we get

$\begin{aligned}\text{Total cases}&=^9C_{2}\\ &=\frac{9!}{(9-2)!2!}\\ &=\frac{9!}{7!2!}\\ &=36\end$

Furthermore, the probability that both are of same colour is

$\begin{aligned}\text{probability}&=\frac{\text{favourable cases}}{\text{Total cases}}\\ &=\frac{10}{36}\\ &=\frac{5}{18}\end$

Hence, none of the choice is correct.

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