There are two boxes. The first box contains 6 green and 4 red balls.

yashsokhal23

Box 1: 6 green + 4 red

Box 2: 3 green + 7 red

Now there are two routes here:
1) Green is drawn from box 1 and then Green is drawn from box 2.
2) Red is drawn from box 1 and then Green is drawn from box 2.

When solving this question one needs to take into account the probability of drawing Green or Red in the first draw AND then the probability of drawing Green from the second draw in either case and then see what the probability is of either the 1st case OR second case.

1) Probability that Green is first drawn and then Green is drawn: \(\frac{6}{10}*\frac{4}{11} = \frac{12}{55}\)


2) Probability that Red is first drawn and then Green is drawn: \(\frac{4}{10}*\frac{3}{11}=\frac{6}{55}\)


Now as there are two possible outcomes, it's an OR case, so we add the two probabilities together:

\(\frac{12}{55}+\frac{6}{55}= \frac{18}{55}\)

Answer B

You have ignored the probability of drawing a red/green from the first box to begin with. Imagine that box one has 9999 green balls and 1 red ball. You'd still have your two possibilities.

1. The ball added from box 1 to 2 was green
2. The ball added from box 1 to 2 was red

In case 1, box 2 has 4 green and 7 red. So probability of drawing green is 4/11.
In case 2, box 2 has 3 green and 8 red. So the probability of drawing green is 3/11.

But we can't just add those up! The odds are 99.99% that we are looking at case 1, so we are FAR more likely to be looking at a case where we have 4/11 probability of drawing green from box 2, so overall probability should basically be 4/11.

This is a two-step probability question. You need to account for the probability of of drawing a green/red from box 1 and apply that to the result from box 2.

Bunuel Could you help us out with this one?

We can have the following two cases:

CASE 1:
If green ball was drawn from the first box and put into the second box, the second box would contain 4 green and 7 red balls. The probability of drawing a green ball from it would be 4/11.

CASE 2:
If red ball was drawn from the first box and put into the second box, the second box would contain 3 green and 8 red balls. The probability of drawing a green ball from it would be 3/11.


The probability of CASE 1, so the probability of drawing a green ball from the first box, is 6/10.
The probability of CASE 2, so the probability of drawing a re ball from the first box, is 4/10.

Therefore, the overall probability of drawing a green ball from the second bo is 6/10*4/11 + 4/10*3/11 = 18/55.

Answer: B.

case 1
green ball drawn from both boxes
6/10 * 4/11
case 2
red ball box 1 and green from box 2
4/10 * 3/11
24+12 / 110 ; 36/110
18/55
option B

the part that you are missing is to give a value to the probability of getting one red or one green in the first part. ( showed in red)

so now probability of getting one green is \(\frac{4}{11} * \frac{6}{10}+\frac{3}{11}*\frac{4}{10}\)
=\(\frac{18}{55}\)