Probability that 2 out of 3 balls will be of the same colour if the first one drawn is red

Question

Three balls are drawn from a box containing 3 red, 4 black and 5 white balls. Find the probability that 2 out of 3 balls will be of the same colour (event A) if the first one drawn is red (event B).

I tried to solve this question using combinations and conditional probability. First, I calculated $P(A \cap B) = \frac{{3\choose1}{4\choose2}+{3\choose1}{5\choose2}+{3\choose2}{4\choose1}+{3\choose2}{5\choose1}} {12\choose3} = 15/44 $ and then I calculated $P(B) = \frac{3\choose1} {12\choose1} = 1/4$. Then I used the two results to find $P(A|B) = \frac{15/44}{1/4} = 1.36$, which is wrong (correct result is $34/55$). I know there are other ways of solving this exercise, but if possible, I would like to solve it using combinations.

Thanks in advance!

Answer 1

It is given that the first ball is red, so in effect you only have 2R,4B, 5W balls left, and only $2$ more to draw

$$ Pr= \frac{\binom21\binom{9}1+\binom42 +\binom 52}{\binom{11}2}= \frac{34}{55} $$

Answer 2

Alternative approach:

The problem may be attacked by dispensing with elegance and very carefully applying brute force.

In general, the number of ways of drawing exactly $3$ balls, with exactly $2$ of the $3$ balls the same color, given that the first ball drawn is red may be computed as follows:

$$\binom{3}{1} \times \left\{ ~\left[\binom{2}{1} \times \binom{4}{1}\right] + \left[\binom{2}{1} \times \binom{5}{1}\right] + \binom{4}{2} + \binom{5}{2} ~\right\}.\tag1 $$

In (1) above, the LHS factor represents the number of ways of initially selecting which red ball will be the first (red) ball drawn.

In (1) above, within the sum of $(4)$ terms, the first term represents the number of ways that the 2nd and 3rd balls drawn will be a red and black ball.

Similarly, the second term represents the number of ways that the 2nd and 3rd balls drawn will be a red and white ball.

Further, the third and fourth terms represent the number of ways that the 2nd and 3rd balls drawn will be two black or two white balls, respectively.

So, the computation in (1) above computes to

$$102.$$

This means that in accordance with Bayes Theorem, you will have to calculate the value $D$ extremely carefully, so that the desired probability may be expressed as

$$\frac{102}{D}.$$

There are two conditions that need to be observed when computing $D$:

• It must be assumed that the first ball drawn is red. This is because (in effect), if you have two events $E_1,E_2$, then by Bayes Theorem, you compute
  
  $\displaystyle p(E_1 | E_2) = \frac{p(E_1,E_2)}{p(E_2)}.$

• In (1) above, no distinct enumerations where given with respect to the order that the 2nd and 3rd balls were drawn. For example, while there are $~\displaystyle \binom{5}{2}~$ ways that the 2nd and 3rd ball can both be white, there are actually $(5 \times 4)$ ways of selecting a white ball as the 2nd ball drawn and then selecting one of the remaining white balls as the 3rd ball drawn.
  
  The point is that when $D$ is enumerated, since there was no distinction made when computing $N$ with respect to the order that the 2nd and 3rd balls were drawn, $D$ must be enumerated in a consistent manner.

Consequently, $D$ may be computed as

$$\binom{3}{1} \times \binom{11}{2} = 165. \tag2 $$

That is, to compute $D$, you reason that there are $(3)$ ways that the first ball drawn may be red. Then, once the first ball drawn is red, you have to enumerate the number of ways of drawing two balls from the remaining $(11)$ balls.

So, the overall probability, using brute force and dispensing with elegance is computed as

$$\frac{102}{165}.$$

Answer 3

First, I calculated $P(A \cap B) = \frac{{3\choose1}{4\choose2}+{3\choose1}{5\choose2}+{3\choose2}{4\choose1}+{3\choose2}{5\choose1}} {12\choose3} = 15/44 $

This is the probability that exactly two among the three are the same colour and at least one among the three is red. That is not the required event.

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Event $B$ specifies the first position's colour. So therefore, you must count ways to fill it separately from the later two in both the numerator and denominator.

The probability that the first one is red and exactly two among the first three are the same colour

$\qquad\begin{align}\mathsf P(A\cap B)&=\dfrac{\dbinom{3}{1}\left[\dbinom{2}{1}\dbinom{4+5}{1}+\dbinom{4}{2}+\dbinom{5}{2}\right]}{\dbinom{12}{1}\dbinom{11}{2}}\end{align}$

And so:

$\qquad\begin{align}\mathsf P(A\mid B)&=\dfrac{\dbinom{2}{1}\dbinom{4+5}{1}+\dbinom{4}{2}+\dbinom{5}{2}}{\dbinom{11}{2}}\end{align}$