# Thread: Law of Total Probability

1. ## Law of Total Probability

One bag contains 6 white and 4 black balls, and a second bag contains 3 white and 7 black balls. One ball is drawn at random from the first bag and placed unseen in the second bag.
(a) Find the probability that a ball now drawn at random from the second bag will be black.
(b) If the ball drawn from the second bag is black, find the probability that a black ball was transferred from the first bag.

I drew a tree diagram for both of them but the fact that the balls get switched into another bag confuses me : / need some help

One bag contains 6 white and 4 black balls, and a second bag contains 3 white and 7 black balls. One ball is drawn at random from the first bag and placed unseen in the second bag.
(a) Find the probability that a ball now drawn at random from the second bag will be black.
(b) If the ball drawn from the second bag is black, find the probability that a black ball was transferred from the first bag.

I drew a tree diagram for both of them but the fact that the balls get switched into another bag confuses me : / need some help

(a) there are 2 cases

1. a white ball was transferred from the 1st bag and a black ball was chosen from the 2nd..
this probability is

$\frac{6}{10}\frac{7}{11}$

since there are now 11 balls in the 2nd bag.

2. a black ball was transferred from the 1st bag and a black ball was chosen from the 2nd...
this probability is

$\frac{4}{10}\frac{8}{11}$

Summing these probabilities gives the probability of choosing a black from
the 2nd bag after a ball was transferred from the 1st.

(b) The probability that a black ball was moved to the 2nd bag is 0.4.

Hence, maybe this question should read...

if a black ball is chosen, what is the probability that it came from the 1st bag?

It has no chance of being from the 1st bag if a white was transferred.

It has a $\frac{1}{8}$ chance of being from the 1st bag if a black was transferred.

Hence, the probability is

$(0)\frac{1}{7}+\frac{4}{10}\frac{1}{8}$