Conditional Probability

[sisxixon]

Bag A contains four yellow marbles and one green marble. Bag B contains two yellow marbles and five green marbles. If a marble is randomly selected from Bag A and placed in Bag B, and a marble is then randomly selected from Bag B, what is the probability that the marble slected from Bag B will be yellow?

can someone please check my work? it's been a while since i've gone over this, and i'd like to make sure:

probability of yellow in bag A: 4/5

probability of yellow in bag B, if the marble drawn from bag A is yellow: (3/8)(4/5)=3/10

probability of yellow in bag B, if the marble drawn from A is green: 2/8

adding both probabilities would equal 11/20, the probability of drawing out yellow from B.

 

[soroban]

Hello, sisxixon!

You did great . . . except for one step.

Bag A contains four yellow marbles and one green marble.
Bag B contains two yellow marbles and five green marbles.
If a marble is randomly selected from Bag A and placed in Bag B,
and a marble is then randomly selected from Bag B,
what is the probability that the marble slected from Bag B will be yellow?

probability of yellow in bag A: \(\displaystyle \,\frac{4}{5}\)

probability of yellow in bag B, if the marble drawn from bag A is yellow: \(\displaystyle \,\left(\frac{3}{8}\right)\left(\frac{4}{5}\right)\,=\,\frac{3}{10}\)

probability of yellow in bag B, if the marble drawn from A is green: \(\displaystyle \,\frac{2}{8}\;\) . . . no

Probability of green from A is \(\displaystyle \frac{1}{5}\).
Probability of yellow in bag B, if green is drawn from A: \(\displaystyle \,\left(\frac{1}{5}\right)\left(\frac{2}{8}\right) \,= \,\frac{1}{20}\)