Question

Box A contains 4 Blue counters and 3 white counters. Box B contains 5 white counters and 3 blue counters. Lauren takes a counter randomly from Box A and put it in Box B. She then takes a counter from Box B and puts it in Box A. WHat is the probability that each box then has the original number of counters of each colour.

Answer 1

Answer: 20/9

Step-by-step explanation:

Answer 2

The probability that each box then has the original number of counters of each color is $\dfrac{34}{63}\rm\;or\;0.5397$.

Given information:

Box A contains 4 Blue counters and 3 white counters. Box B contains 5 white counters and 3 blue counters.

Lauren takes a counter randomly from Box A and puts it in Box B. She then takes a counter from Box B and puts it in Box A.

It is required to calculate the probability that each box then has the original number of counters of each color.

This can happen in two cases:

1. She draws a blue counter from box A and puts it in box B, and then draws a blue counter from box B and puts it in box A. Or,
2. She draws a white counter from box A and puts it in box B, and then draws a white counter from box B and puts it in box A.

If she draws a blue counter from box A, then the total probability of the event will be,

$P(B)=\dfrac{4}{7}\times \dfrac{4}{9}\\=\dfrac{16}{63}$

If she draws a white counter from box A, then the total probability of the event will be,

$P(W)=\dfrac{3}{7}\times \dfrac{6}{9}\\=\dfrac{2}{7}$

The total probability that each box will have the same configuration as the initial will be,

$P=P(B)+P(W)\\P=\dfrac{16}{63}+\dfrac{2}{7}\\P=\dfrac{34}{63}\\P=0.5397$

Therefore, the probability that each box then has the original number of counters of each color is $\dfrac{34}{63}\rm\;or\;0.5397$.

For more details, refer to the link:

https://brainly.com/question/795909