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?

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

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

Blue chip transferred, red coin chosen:

6/10 of 5/9. So

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?

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

b

2

−

b

b

3

+

7

b

2

+

14

b

+

6

%

−

3

b

3

+

7

b

2

+

14

b

+

6

%

step-by-step explanation:

c.

step-by-step explanation:

she should of added 9^2 and 12^2. that is because it is a^2+b^2=c^2. c^2 is the hypotenuse and should always be bigger than the other two numbers. which is why she should of added 9 squared and 12 squared together instead of 15 squared.