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?

Step-by-step explanation:

Given that 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.

The poker chip taken can be red with p = 4/9 or blue with prob 5/9

If red is taken, box B would have 7 red and 3 blue or if blue is taken, box B would have 6 red and 4 blue.

Let A1 - first ball drawn is blue

A2 - first ball drawn is red

B - ball drawn from second box is red

Required probability =

P(A1B) =

P(A2B) = " alt="\frac{\frac{2}{9}}{\frac{2}{9}+\frac{14}{45}} \\=\frac{10}{24} \\=\frac{5}{12}" />" />

substitute to get required probability

=

it's

i believe the answer is the polygon, heptagon.