Question

From a bag containing 10 distinct balls, 6 balls are drawn simultaneously and replaced. Then 4 balls are drawn. The probability that exactly 3 balls are common to the drawings is

• A. $\frac{8}{21}$
• B. $\frac{6}{19}$
• C. $\frac{5}{24}$
• D. $\frac{9}{22}$

Answer 1

The correct option is A $\frac{8}{21}$

Let S be the sample space of the composite experiment of drawing 6 in the first draw and then four in second draw then $n\left(S\right){=}^{10}{C}_6{\times }^{10}{C}_4$
We will first select the 6 balls from 10 balls. Next step will be selecting 3 balls common for the round of drawing six balls and the round of drawing 4 balls.We have to select three balls from the first six balls. After that we have to select the last ball from the remaining 4 balls.
$\therefore$ Required Probability $=\frac{{}^{10}{C}_6{\times }^6{C}_3{\times }^4{C}_1}{{}^{10}{C}_6{\times }^{10}{C}_4}=\frac{80\times 24}{10\times 9\times 8\times 7}=\frac{8}{21}$