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Suppose I have two boxes. They look identical to me. However, I am told: “Box A contains 1 red ball and three white balls; Box B has 2 red balls and 2 white balls.”

I randomly pick a box and select a ball from the box without looking into the box.

It is a red ball. What is the probability that I have picked Box A? Choose one of the following options.

A. Less than 30%. B. 30% ~ 50% (30 % is included; 50% is not included). C 50%. D. 50% ~ 60% (50% is not included; 60% is included). E. More than 60%.

Solution

P(Picking Red ball)$=(1+2)/(1+3+2+2)=3/8$

P(Picking Red Ball From Box A)$=1/2*1/4=1/8$

We want P(Picking Box A|Red Ball Picked)=P(Picking a Red ball from Box A)\P(Red Ball Picked)

Therefore, probability is $=(1/8)/(3/8)=1/3$ Answer is A.

Am I doing this correctly? My tutor is known for giving not-so straightforward questions, so I'm wondering if I need to consider another way, or I could be wrong. Any alternatives welcome too!

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2 Answers 2

up vote 2 down vote accepted

$P\text{(picking a red ball)} =\frac 12 (\text {pick A}) \frac 14 (\text{\pick red ball (from A))}\\+ \frac 12 (\text {pick B}) \frac 12 (\text{\pick red ball (from B))}=\frac 18 + \frac 14=\frac 38$

This agrees with your value, but only because there were the same number of balls in each box, so each ball is equally likely to be picked. If box B had 2 red and 4 white the probability of a red ball would be $\frac 12\frac 14 + \frac 12 \frac 13=\frac 18 + \frac 16=\frac 7{24}$ , but you would have $\frac {1+2}{4+6}=\frac 3{10}$

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You calculated the probability correctly, but the answer is B, not A: $\frac13$ is not less than $30$%.

You could also have argued, less formally but just as correctly, that there are $8$ equally likely outcomes, and you’re having picked a red ball eliminates $5$ of them. Of the remaining $3$, only one involves picking a ball from box $A$, so the probability that you picked from box $A$ is $\frac13$.

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Thanks, @Ross.$\;$ –  Brian M. Scott Oct 12 '12 at 14:21

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