Problem 5 - Set 9 - Try Out - Quantitative Knowledge - SNBT

Number 5

Inside a box there are $1$ red ball, $6$ green balls, and $2$ white balls. If $7$ balls are to be drawn without replacement, then the probability that the number of green balls drawn is twice the number of white balls drawn is...

$\frac{1}{12}$

$\frac{1}{4}$

$\frac{5}{12}$

$\frac{1}{2}$

$\frac{2}{3}$

Explanation

Let $A$ be the event that the number of green balls drawn is twice the number of white balls drawn.

Given:

There are $9$ balls in a box ( $1$ red + $6$ green + $2$ white = $9$ ). $7$ balls will be drawn at random. The total number of ways to draw the balls is:

n(S) = C^9_7 = \frac{9!}{7!(9-7)!} = \frac{9 \times 8}{2 \times 1} = 36

We want to find the probability that the number of green balls is twice the number of white balls. We divide this into two cases.

Case 1

If $1$ white ball is drawn, then there must be $2$ green balls (since green = 2 times white). To reach a total of $7$ balls drawn, the remaining balls to be drawn must be $7 - 1 - 2 = 4$ balls. These must be red balls. However, there is only $1$ red ball available. Therefore, Case 1 is impossible.

Case $2$

If $2$ white balls are drawn, then there must be $4$ green balls. To reach a total of $7$ balls drawn, the remaining balls to be drawn must be $7 - 2 - 4 = 1$ ball. This must be a red ball. This case is possible because there are $2$ white balls, $6$ green balls, and $1$ red ball available.

The number of ways to draw for this case is:

n(A) = C^2_2 \cdot C^6_4 \cdot C^1_1

= 1 \cdot 15 \cdot 1

= 15

Conclusion

Thus, the probability of event $A$ is:

P(A) = \frac{n(A)}{n(S)} = \frac{15}{36} = \frac{5}{12}

Therefore, the probability that the number of green balls drawn is twice the number of white balls drawn is $\frac{5}{12}$ .

Command Palette

Number 5

Explanation