# Thread: probability of colours of balls

1. ## probability of colours of balls

the question is :

(b) Box 1 contains six red balls and four green balls and box 2 contains seven red balls and three green balls. A ball is chosen at random from box 1 and placed in box 2. A ball is then chosen at random from box 2 and placed in box 1.

Calculate
(i) the probability that a red ball is selected from both box 1 and box 2;

(ii) the probability that, at the conclusion of the selection process, there are again six red balls and four green balls in box 1.

thanks

2. Originally Posted by master_2m8
the question is :

(b) Box 1 contains six red balls and four green balls and box 2 contains seven red balls and three green balls. A ball is chosen at random from box 1 and placed in box 2. A ball is then chosen at random from box 2 and placed in box 1.

Calculate
(i) the probability that a red ball is selected from both box 1 and box 2;

(ii) the probability that, at the conclusion of the selection process, there are again six red balls and four green balls in box 1.

thanks
Hi,

i) $\displaystyle P(\text{red in box 1})=\frac{6}{10}$
I assume that the red ball is placed in box 2 now:
$\displaystyle P(\text{red in box 2})=\frac{8}{11}$
$\displaystyle P(\text{red from box 1 and 2})=(\frac{6}{10})(\frac{8}{11})$

ii) We can look at two cases when red ball is transfered and green ball is transfered:
$\displaystyle P(\text{6 red and 4 green in box 1 after transfer})= (\frac{6}{10})(\frac{8}{11})+(\frac{4}{10})(\frac{ 4}{11})$

3. ## colour of ball

a) Pr( red in the first time and red in the second) = Pr(red in the first). Pr(red in the second | red in the first) (conditional probability)

= 6/10 . 8/11 = 24/55
b) A: the event red in the first and red in the second moving .
Ag: the event green in the first moving and green in the second
B: the event that after 2 moving there are again 6 red and 4 green in the box 1. We need calculate Pr(A|B) = ?
Pr(A|B) = Pr(AB)/ Pr(B) = Pr(A)/Pr(B) (because A \subset B)

We calculate Pr(B):
Pr(B) = Pr(A) + Pr(Ag) --> the result