Question

a box p contains 3 white and 4 green balls. another box q contains 5 white and 2 green balls. a ball is drawn at random from box p and dropped into box Q. a ball is then drawn at random from box Q. find the probability that the ball drawn from box Q is white.

Answer 1

For the first part of experiment, we have two outcomes,

A = white ball with probability 3/7

We then have 6 white and 2 non-white balls.

P(W)=3/7*6/8=9/28

B= non-white ball with probability 4/7

We then have 5 white and 3 non-white balls

P(W)=4/7*5/8 = 5/14

Total probability of picking a white ball in the second step = 9/28+5/14=19/28

Answer 2

p: 3 white 4 green → white drawn: 3 out of 7 ($\frac{3}{7}$)

→ green drawn: 4 out of 7 ($\frac{4}{7}$)

q: 5 white 2 green → white drawn from p: 6 out of 8 ($\frac{3}{4}$)

→ green drawn from p: 5 out of 8 ($\frac{5}{8}$)

Probability of white from p and white from q:

$\frac{3}{7}$ x $\frac{3}{4}$ = $\frac{9}{28}$

Probability of green from p and white from q:

$\frac{4}{7}$ x $\frac{5}{8}$ = $\frac{5}{14}$

white p & white q or green p & white q

$\frac{9}{28}$ or $\frac{5}{14}$

$\frac{9}{28}$ or $\frac{10}{28}$ = [tex] \frac{19}{28}

Answer: [tex] \frac{19}{28}