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Maslowich
2 years ago
11

Suppose box A contains 4 red and 5 blue poker chips and box B contains 6 red and 3 blue poker chips. Then a poker chip is chosen

at random from box A and placed in box B. Now, a poker chip is chosen at random from those now in box B. 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?
Mathematics
1 answer:
sergejj [24]2 years ago
5 0

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)}

In which

P(B|A) is the probability of event B happening, given that A happened.

P(A \cap B) 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

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

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

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

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