Question

A bag contains 2 yellow, 3 black and 2 green balls. Two balls are drawn at random. What
is the probability that none of the balls drawn is green?
11/21
12/21
10/21
none of the above​

Answer 1

Step-by-step explanation:

sorry there is some mistake in question justify it

Answer 2

Answer:

The probability that none of the balls drawn is green is $\frac{10}{21}$.

Step-by-step explanation:

Total number of ball=2+3+2= 7 balls

As two balls are drawn at random, the total number of possible

chances =$\begin{gathered}{ }^{7} C_{2} \end{gathered}$

Total number of balls other than green=5 balls

So, the number of chances that none of the balls drawn is green=$\begin{gathered}{ }^{5} C_{2} \end{gathered}$

Therefore, the probability that none of the balls

drawn is green=   $\frac{{ }^{7} C_{2}}{{ }^{5} C_{2}}$

                          =$\frac{5 \times 4}{7 \times 2}$

                          =$\frac{10}{21}$