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!
Best Answer
$P\text{(picking a red ball)} =\frac 12 (\text {pick A}) \frac 14 (\text{\pick red ball (from A))}\\+ \frac 12 (\text {pick B}) \frac 12 (\text{\pick red ball (from B))}=\frac 18 + \frac 14=\frac 38$
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 $\frac {1+2}{4+6}=\frac 3{10}$