Answer:
0.5172 = 51.72% probability a blue poker chip was transferred from box A to box B, given that the coin just chosen from box B is red
Step-by-step explanation:
Conditional Probability
We use the conditional probability formula to solve this question. It is
![P(B|A) = \frac{P(A \cap B)}{P(A)}](https://tex.z-dn.net/?f=P%28B%7CA%29%20%3D%20%5Cfrac%7BP%28A%20%5Ccap%20B%29%7D%7BP%28A%29%7D)
In which
P(B|A) is the probability of event B happening, given that A happened.
is the probability of both A and B happening.
P(A) is the probability of A happening.
In this question:
Event A: Coin chosen from box B is red.
Event B: Blue poker chip transferred.
Probability of choosing a red coin:
7/10 of 4/9(red coin from box A)
6/10 of 5/9(blue coin from box A). So
![P(A) = \frac{7}{10}*\frac{4}{9} + \frac{6}{10}*\frac{5}{9} = \frac{28 + 30}{90} = 0.6444](https://tex.z-dn.net/?f=P%28A%29%20%3D%20%5Cfrac%7B7%7D%7B10%7D%2A%5Cfrac%7B4%7D%7B9%7D%20%2B%20%5Cfrac%7B6%7D%7B10%7D%2A%5Cfrac%7B5%7D%7B9%7D%20%3D%20%5Cfrac%7B28%20%2B%2030%7D%7B90%7D%20%3D%200.6444)
Blue chip transferred, red coin chosen:
6/10 of 5/9. So
![P(A \cap B) = \frac{6}{10}*\frac{5}{9} = \frac{30}{90} = 0.3333](https://tex.z-dn.net/?f=P%28A%20%5Ccap%20B%29%20%3D%20%5Cfrac%7B6%7D%7B10%7D%2A%5Cfrac%7B5%7D%7B9%7D%20%3D%20%5Cfrac%7B30%7D%7B90%7D%20%3D%200.3333)
What is the probability a blue poker chip was transferred from box A to box B, given that the coin just chosen from box B is red?
![P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.3333}{0.6444} = 0.5172](https://tex.z-dn.net/?f=P%28B%7CA%29%20%3D%20%5Cfrac%7BP%28A%20%5Ccap%20B%29%7D%7BP%28A%29%7D%20%3D%20%5Cfrac%7B0.3333%7D%7B0.6444%7D%20%3D%200.5172)
0.5172 = 51.72% probability a blue poker chip was transferred from box A to box B, given that the coin just chosen from box B is red