Red and black!

Probability Level 4

Box $A$ contain 6 red and 4 black balls, and box $B$ contains 4 red and 6 black balls. One ball is drawn at random from box $A$ and placed in box $B$ , and then one ball is taken from box $B$ and placed in box $A$ . If one ball is now drawn at random from box $A$ , then find the probability that it is red.

If the probability is in the form $\frac{a}{b}$ , where $a$ and $b$ are coprime positive integers, submit your answer as $a+b$ .

The answer is 87.

2 solutions

Chew-Seong Cheong
Mar 23, 2017

Consider the expected numbers of red and black balls of the stages.

Stage 0: Originally

$\begin{array} {rcc} & \color{#D61F06} \text{Red} & \text{Black} & \color{#3D99F6} \text{Total} \\ \color{#3D99F6} \text{Box }A & \color{#D61F06} 6 & 4 & \color{#3D99F6} 10 \\ \color{#3D99F6} \text{Box }B & \color{#D61F06} 4 & 6 & \color{#3D99F6} 10 \end{array}$

Stage 1: One ball is drawn at random from box $A$ and placed in box $B$ .

$\begin{array} {rcc} & \color{#D61F06} \text{Red} & \text{Black} & \color{#3D99F6} \text{Total} \\ \color{#3D99F6} \text{Box }A & \color{#D61F06} 6 - \frac 6{10} = \frac {27}5 & 4 - \frac 4{10} = \frac {18}5 & \color{#3D99F6} 9 \\ \color{#3D99F6} \text{Box }B & \color{#D61F06} 4 + \frac 6{10} = \frac {23}5 & 6 + \frac 4{10} = \frac {32}5 & \color{#3D99F6} 11 \end{array}$

Stage 2: Then one ball is taken from box $B$ and placed in box $A$ .

$\begin{array} {rcc} & \color{#D61F06} \text{Red} & \text{Black} & \color{#3D99F6} \text{Total} \\ \color{#3D99F6} \text{Box }A & \color{#D61F06} \dfrac {27}5 + \dfrac {\frac {23}5}{11} = \dfrac {64}{11} & \dfrac{18}5 + \dfrac {\frac {32}5}{11} = \dfrac {46}{11} & \color{#3D99F6} 10 \\ \color{#3D99F6} \text{Box }B & \color{#D61F06} \dfrac {23}5 - \dfrac {\frac {23}5}{11} = \dfrac {46}{11} & \dfrac {32}5 - \dfrac {\frac {32}5}{11} = \dfrac {64}{11} & \color{#3D99F6} 10 \end{array}$

After stage 2, the probability of drawing a red ball from box $A$ is $\dfrac {\frac {64}{11}}{10} = \dfrac {32}{55}$ .

$\implies a+b = 32+55 = \boxed{87}$

I'm wondering if anybody could comment as to the flaw of reasoning of the following "solution". It must be incorrect, but I don't see where I am going wrong. There are four cases: Case 1: (R,R). We choose a red ball from box A and then a red ball from box B. The probability of this happening would be (6/10) (5/11) (because, after we put a red ball into box B, it would then have 5 red balls and 6 black balls in box B). If we choose (R,R), then box A is left with 6 red balls and 4 black. Which means, the Probability of choosing a red ball given we have chosen (R, R) is 6/10. Case 2: (R, B). By a similar reasoning as above, the probability of making that choice is (6/10) (6/11). Given that we have chosen (R, B), box A now has 5 red balls and 5 black. So, Prob(choosing Red given (R,B)) = 5/10. Case 3: (B, R). The probability of this choice is (4/10) (4/11). We would then have 7 red balls and 3 black in box A, so Prob.(choosing Red given (B,R)) = 7/10. Lastly, Case 4 (B, B). The probability of this happening is (4/10) (7/11). In this case, we are left with 6 red and 4 black in box A, so that P(choosing red given (B, B)) = 6/10. Adding up across all these cases would seem to imply that Prob(choosing red) = Prob(red given (R,R))*P(R,R) + ... + Prob(red given (B,B)) = 584/1100 = 146/275. I'm fairly certain my arithmetic is correct (I've checked it several times), so the reasoning must be flawed. I'm not sure why. Any comments would be greatly appreciated. Thanks in advance.

Ron Gallagher - 1 year, 2 months ago

Zee Ell
Mar 22, 2017

Let's call the combination of red and black balls:

• in box A, after taking a ball from box A and putting into box B: A'

• in box B, after taking a ball from box A and putting into box B: B'

(e.g. B' (4R, 7B) means, that we have 4 red and 7 black balls in box B at this point)

• in box A, after taking a ball from box B' and putting into box A: A''

We start with the following two boxes:

A°(6R, 4B) and B°(4R, 6B)

Now consider the following 2 × 2 = 4 cases:

I. The first ball (taken from A°, put into B°) is red:

$p(A'(5R, 4B)) = p(B'(5R, 6B)) = \frac {6}{10}$

Case 1: The second ball (taken from B', put into A') is red:

$p(A''(6R, 4B)) = \frac {6}{10} × \frac {5}{11} = \frac {30}{110}$

Case 2: The second ball (taken from B', put into A') is black:

$p(A''(5R, 5B)) = \frac {6}{10} × \frac {6}{11} = \frac {36}{110}$

II. The first ball (taken from A°, put into B°) is black:

$p(A'(6R, 3B)) = p(B'(4R, 7B)) = \frac {4}{10}$

Case 3: The second ball (taken from B', put into A') is red:

$p(A''(7R, 3B)) = \frac {4}{10} × \frac {4}{11} = \frac {16}{110}$

Case 4: The second ball (taken from B', put into A') is black:

$p(A''(6R, 4B)) = \frac {4}{10} × \frac {7}{11} = \frac {28}{110}$

Therefore, the probability of getting a red ball (from box A) at the third draw:

$\frac {6}{10} × \frac {30}{110} + \frac {5}{10} × \frac {36}{110} + \frac {7}{10} × \frac {16}{110} +\frac {6}{10} × \frac {28}{110} = \frac {640}{1100} = \frac {32}{55}$

$\text {Hence, a = 32, b = 55 and our answer should be: }$

$a + b = 32 + 55 = \boxed { 87 }$

Remark:

Alternatively, we could have used either a probability tree diagram or a table to show the outcomes and their probabilities.

Red and black!

The answer is 87.

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