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An urn contains 6 white and 9 black balls. If 4 balls are to be randomly selected without replacement, what is the probability that the first 2 selected are white and the last 2 black.

I solved it in this manner : (6C2 * 9C2)/ (15C4)    --- getting 36/91

but the answer is (6*5*9*8)/(15*14*13*12)   --- 6/91

I cannot understand exactly why my approach is not working over here.

You are calculating probability of is choosing 2 white balls and 2 black balls without any order.

Number of white balls = 6

Number of black balls = 9

Probability of 1st ball being white = $\large \frac{6}{15}$

Probability of 2st ball being white = $\large \frac{4}{14}$

Probability of 3rd ball being black = $\large \frac{9}{13}$

Probability of 4th ball being black = $\large \frac{8}{12}$

Probability = $\large \frac{6}{15} * \frac{4}{14} * \frac{9}{13} * \frac{8}{12} = \frac{6}{91}.$

### 1 comment

This question is given in the conditional probability exercise. Can you explain to me solve this problem using conditional probability?

Let the event that the

$\rightarrow1^{st}$ is drawn is white be $W_{1}$

$\rightarrow2^{nd}$  ball is white be $W_{2}$

$\rightarrow3^{rd}$ ball is black be $B_{3}$

$\rightarrow4^{th}$  ball is black be $B_{4}$.

$P(W_{1}\cap W_{2}\cap B_{3}\cap B_{4}) \$$\rightarrow P(W_{1}).P(W_{2}/W_{1}).P(B_{3}/(W_{1}\cap W_{2}).P(B_{4}/(W_{1}\cap W_{2}\cap B_{1}))$

$\rightarrow (6/15).(5/14).(9/13).(8/12)$

$\rightarrow 6/91$

Or

simply using multiplication principle and drawing one ball at a time will give

the probability of drawing $1^{st}$  ball which is white  is $6/15$

the probability of drawing $2^{nd}$  ball which is white is $5/14$

the probability of drawing $3^{rd}$ ball which is black  is  $9/13$

the probability of drawing $4^{th}$  ball which is black is $8/12$

$\rightarrow (6/15).(5/14).(9/13).(8/12)$

$\rightarrow 6/91$

Your answer should be $\binom{6}{2}*\binom{9}{2}/\binom{15}{2}*\binom{13}{2}$        $\because$ balls are to be randomly selected without replacement