Homework: What is the probability that I have picked Box A? 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!
 A: $P\text{(picking a red ball)} =\frac 12 (\text {pick A}) \frac 14 (\text{\pick red ball (from A))}\\+ \frac 12  (\text {pick B}) \frac 12 (\text{\pick red ball (from B))}=\frac 18 + \frac 14=\frac 38$  
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 $\frac {1+2}{4+6}=\frac 3{10}$
A: 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 your 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 $\frac13$.
A: 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
Let P(A) be Box A Let P(B) be Red Ball
P(AB) = P(A|B) P(B) or P(B|A) P(A)
Since we want Box A | Red Ball, so we pick, A(AB) = P(A|B) P(B)
Based on A(AB) = P(A|B) P(B), We can conclude that, P(A|B) = A(AB) / P(B)
P(A|B) = 1/8 / 3/8 = 1/3
Therefore, Answer is B. >30%, (1/3) or 33.3%
You have done it perfectly correct but marked the wrong option.
so the answer is 33.333%.
A: Given that probability of choosing any box is equal.
→p(boxA│red ball)=$\frac{p(red ball|boxA)p(boxA)}{p(red ball)}$
→p(boxA│red ball)= $\frac{p(red ball|boxA)p(boxA)}{(p(red ball|boxA)p(boxA)+p(red ball|boxB)p(boxB))}$
→p(boxA│red ball)=$\frac{(\frac14×\frac12)}{(\frac12  \frac14+\frac24 \frac12)}$
→p(boxA│red ball)=$\frac{(\frac18)}{(\frac18+\frac28)}$=1/3
A: 
When you randomly pick a box so it is A or either B so probability would be 0.5
