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A box contains six red balls and four green balls. Four balls are selected at random from the box. What is the probability that two of the selected balls will be red and two will be in green?

1. $\frac{1}{14}$
2. $\frac{3}{7}$
3. $\frac{1}{35}$
4. $\frac{1}{9}$

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## 1 Answer

+1 vote
4 balls can be selected from 10 balls in ${10_{C}}_{4}$ ways.

2 red balls can be selected from 6 red balls in ${6_{C}}_{2}$ ways.

2 green balls can be selected from 4 green balls in ${4_{C}}_{2}$ ways.

hence probability= (${6_{C}}_{2}$ *${4_{C}}_{2}$)/${10_{C}}_{4}$ = 3/7

Answer: (B)
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