# Probability of drawing a ball in sequence without replacing

## My Question

$\text{In a box,there are$2$red,$3$black and$4$blue coloured balls. The probability of drawing }$ $2\text{ blue balls in sequence without replacing and then drawing$1$black ball from this box is}$

## My Approach

Reqd probability$$=\frac{\binom{4}{2} \binom{3}{1}}{\binom{9}{3}}$$

$$=\frac{3}{14}=21.4 \%$$ but the answer is given as $\text{6.80 to 7.20(Range)} \%$ Am i doing it right ?Please help me out. Thanks

• The order is specified. So: $\frac 4{9}\times \frac 38\times \frac 37$. Your computation gives the probability of drawing $2$ blue and $1$ black in any order, so three times the correct answer.
– lulu
Commented May 14, 2018 at 12:22
• We could modify your approach: The probability that the first two balls are blue is $\frac{\binom{4}{2}}{\binom{9}{2}}$. The probability that the third ball selected is black given that two blue balls have already been selected is $\frac{\binom{3}{1}}{\binom{7}{1}}$ since three of the seven balls remaining are black. Hence, the required probability is $$\Pr(\text{two blue, then one black}) = \frac{\binom{4}{2}}{\binom{7}{2}} \cdot \frac{\binom{3}{1}}{\binom{7}{1}} = \frac{1}{14}$$ Commented May 14, 2018 at 12:49
• @N.F.Taussig thanks a lot.This was the point i was looking for ! Commented May 15, 2018 at 5:14

$$\frac{4}{9} \cdot \frac{3}{8} \cdot \frac{3}{7}$$
• @laura No ... you have calculated the probability of picking 2 blue and 1 black balls as a set out of the $9$ ... which is equivalent to picking $3$ balls without replacement ... but that would be compatible with first drawing a black and then 2 blue ones. But it is specified that the third one should be black. Also, if the problem was stated as If it was with replacement, it would be $\frac{4}{9} \cdot \frac{4}{9} \cdot \frac{3}{9}$ Commented May 14, 2018 at 12:25