Three balls are drawn at random without replacement what is the probability that first ball is black second ball is black and third ball is white

I got two answer and confused which one is right or no one is right

first is we can select first two black balls by 10C2 ways and white balls by 20 ways so total no of ways of selecting an event is 10C2*20. And total no of ways to select 3 balls is 30C3. ans is 10C2*20/30C3.

and second approach was to select one black ball the prob is 10/30 followed by to select another black ball the prob is 9/29 and to select white ball is 20/28 Ans is 10*9*20/30*29*28

  • 2
    $\begingroup$ The second approach gives the correct answer. The denominator in the first approach ${30\choose 3}$ suggests that the balls can come out of the urn in any order, and not specifically BBW. $\endgroup$ – Doug M Nov 30 '19 at 8:07
  • $\begingroup$ Your first approach ignores that the order of the balls matters here. Multiply the total by $3!$ or just use $^30P_3$ (read as 30 permute 3). For the top, your choices are 1 black, another black, then a white, so $10 * 9 * 20$ $\endgroup$ – Varad Mahashabde Dec 3 '19 at 21:08

Your first is a correct answer to another question: "What is the probability that $2$ will be black and $1$ white?"

You can make it an answer to the original question by modification: divide by $3$ because there are $3$ different orders: BBW, BWB, WBB. This results in:$$\frac13\frac{\binom{10}2\binom{20}1}{\binom{30}3}=\frac{15}{203}$$

The second answer is correct. If $B_i$ denotes the event that the $i$-th ball drawn is black for $i=1,2$ and $W_3$ the event that the third ball drawn is white then a formalization of the answer is:$$P(B_1\cap B_2\cap W_3)=P(B_1)P(B_2\mid B_1)P(W_3\mid B_1\cap B_2)=\frac{10}{30}\frac{9}{29}\frac{20}{28}=\frac{15}{203}$$


The second option, $\frac{10\cdot9\cdot20}{30\cdot29\cdot28}$, is correct.


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