There are two bags with red and green balls only, the first bag contai

Question

There are two bags with red and green balls only, the first bag contains 4 red and 6 green balls, second contains 3 red and 7 green balls. Andy picked 1 ball from the first bag and placed it in the second bag, then Mira picked a ball from the second bag. What is the probability that Mira picked a red ball?

A) 8/55
B) 17/55
C) 9/55
D) 1/3
E) 23/110

This was my work:

First bag P(Red) * second bag P(Red) = 4/10*4/11 = 16/110 = 8/55
First bag P(Green) * second bag P(Red) = 6/10*3/11 = 18/110 = 9/55

Answer = 8/55 + 9/55 = 17/55

alldaytestprep · Answer

The probability of picking a red from second bag depends on whether or not a red or green is moved from the first bag. There is a 4/10 chance a red is picked from first bag and a 6/10 chance a green is picked.

If a red is picked from first bag there is a 4/11 chance of picking a red from second bag.

If a green is picked from first bag there is a 3/11 chance of picking a red from second bag

Therefore there is a 40% chance of having a 4/11 chance of picking a red OR there is a 60% chance of having a 3/11 chance of picking a red

(4/10)(4/11) + (6/10)(3/11) = (16/110) + (18/110) = 34/110 = 17/55

Answer: B

pandeyashwin · Answer

There are two possbilities : 
1. Andy picks up a red ball and transfers it to the second bag.
2. Andy picks up a green ball and transfers it to the second bag.

Probability of Andy picking up a red ball and then Mira picking up a red ball = 4/10 * 4/11 (4/11 because Andy transfers the red ball to the second bag)

Probability of Andy picking up a green ball and then Mira picking up a red ball = 6/10 * 3/11 (Total ball increases but not the red ball because Andy transferred a green ball)

Probability = 4/10*4/11 + 6/10*3/11 = 34/110 = 17/55

KarishmaB · Answer

The question may look complicated but it is not. There are two events that take place in sequence. The probability of the second event depends on the result of the first event. We just need to consider the two results of the first event separately. It is like a partial binomial tree. BT.jpeg Probability of picking a Red from second bag = (4/10)*(4/11) + (6/10)*(3/11) = 17/55

bumpbot · Answer

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There are two bags with red and green balls only, the first bag contai

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