# Box A contains 10 red and 5 yellow balls while box B contains 5 red and 6 yellow balls.

Box A contains 10 red and 5 yellow balls while box B contains 5 red and 6 yellow balls. A box is chosen at random and a ball is selected from it. The probability of choosing box A is 55%. Find the probability that the ball came from box B if the ball was yellow.

I'm having a hard time with these questions as this wasn't covered in our lesson. :(

• Have you tried it, what was your attempt? Apr 10 '17 at 1:41
• I wasn't sure how to plug things in. In this lesson the only formula we were given was P(A|B) = P(A & B)/P(B) and I don't know how to plug in the probability of which box. :c Apr 10 '17 at 1:44
• I didn't know how to type in the intersection symbol lol Apr 10 '17 at 1:44
• Here is a tutorial on how to typeset mathematics on this site. Apr 10 '17 at 2:09

Hint: Try to use Bayes' Theorem, where $$P\left(\text{box B}|\text{Yellow}\right)=\frac{P\left(\text{box B}\right)\cdot P\left(\text{Yellow}|\text{Box B}\right)}{P\left(\text{Yellow}\right)}$$ and try to calculate the probabilities on the right hand side based on the given conditions.

• Thank you! I will try using this. :) I appreciate the direction. Apr 10 '17 at 1:48
• I got the answer 0.5724 Is this correct? Apr 10 '17 at 2:16
• It's correct :) Apr 10 '17 at 2:18
• Yay! Thanks for your help. ^.^ Apr 10 '17 at 2:20
• No problem. Glad to help! Apr 10 '17 at 2:20

If we chose from box $A$, and that has $0.55$ probability, then the probability we got a yellow marble is $\frac{5}{15}$. However if we chose from box $B$, and that has $0.45$ probability, then the probability we got a yellow is $\frac{6}{11}$. Hence the probably of getting a yellow by the law of total probability is,

$$(0.55)(\frac{5}{15})+(0.45)( \frac{6}{11})$$

$$\frac{(0.45)( \frac{6}{11})}{(0.55)(\frac{5}{15})+(0.45)( \frac{6}{11})}$$