# Probability of selecting two white balls from two bags

• vcsharp2003
In summary, the conversation discusses the process of identifying and calculating the probability of selecting two white balls from three bags. It mentions three independent events - selecting two bags, selecting a ball from one bag, and selecting a ball from the other bag. The probability is calculated for three mutually exclusive cases and then added together to get the final probability. The correct probability for selecting two bags in each case is determined to be ##\frac {1} {{}^3 C_2} = \frac {1}{3}##, resulting in a final answer of ##\frac {29}{72}##.
vcsharp2003
Homework Statement
Three bags contain 2 white and 1 black balls, 3 white and 3 black balls, 6 white and 2 black balls.
Two bags are selected and a ball is drawn from each. Find the probability (a) that both balls are white and (b) that both balls are of same color.
Relevant Equations
P(A and B and C) = P(A) P(B) P(C), if A,B,C are independent events
P(A or B or C) = P(A) + P(B) + P(C), if A,B,C are mutually exclusive events
My attempt for part (a) is as given below. I will attempt part (b) after getting part (a) correct.

(a) Based on what is asked, we can identify 3 independent events as follows: (i) select any 2 bags followed by (ii) select a ball from one bag followed by (iii) select a ball from the other bag. Also, we will have 3 mutually exclusive events of selecting any 2 of 3 bags, which are listed as Case 1, Case 2 and Case 3. We find the probability of selecting 2 white balls under each case and then add them up to get probability of selecting a white ball from any of the 2 bags.

Case 1: (2w,1b) and (3w,3b)
Probability of selecting these two bags out of three bags ## =\frac {2} {3}##
$$\text{P(2w1)} = \frac {2}{3} \times \frac {3}{6} \times \frac {2}{3} = \frac {2}{9}$$

Case 2: (2w,1b) and (6w,2b)
Probability of selecting these two bags out of three bags ## =\frac {2} {3}##
$$\text{P(2w2) }= \frac {2}{3} \times \frac {6}{8} \times \frac {2}{3} = \frac {1}{3}$$

Case 3: (3w,3b) and (6w,2b)
Probability of selecting these two bags out of three bags = ##\frac {2} {3}##
$$\text{P(2w3) }= \frac {3}{6} \times \frac {6}{8} \times \frac {2}{3} = \frac {1}{4}$$
$$\therefore \text{P(2w) } = \text{P(2w1) } + \text{P(2w2) } + \text{P(2w3) } = \frac {2}{9} + \frac {1}{3} + \frac {1}{4} = \frac {29}{36}$$
But the answer given in the book for part (a) is ##\frac {29} {72}##.
I am not sure where I have made a mistake.

Last edited:
You have 3 disjoint cases for picking the two bags and say that each one has 2/3 probability. That adds up to a lot of probability. You should rethink that.

vcsharp2003
FactChecker said:
You have 3 disjoint cases for picking the two bags and say that each one has 2/3 probability. That adds up to a lot of probability. You should rethink that.
Yes, I made a silly mistake. Thankyou for pointing out my mistake.

The correct way of determining probability of selecting two bags in Case 1 or Case 2 or Case 3 should be ##\frac {1} {{}^3 C_2} = \frac {1}{3}##. Then the answer comes out to be ##\frac {29}{72}##, which matches the answer given in the book.

FactChecker

## 1. What is the probability of selecting two white balls from two bags if each bag contains 5 white balls and 5 black balls?

The probability of selecting two white balls from two bags in this scenario is 1/2 or 50%. This is because there are a total of 10 white balls and 10 black balls, making the total number of possible outcomes 20. Therefore, the probability of selecting two white balls is 10/20 or 1/2.

## 2. What is the probability of selecting two white balls from two bags if one bag contains 4 white balls and the other bag contains 6 white balls?

The probability of selecting two white balls from two bags in this scenario is 2/3 or approximately 66.7%. This is because there are a total of 10 white balls and 10 black balls, making the total number of possible outcomes 20. However, since one bag has more white balls, there are more ways to select two white balls, making the probability 2/3.

## 3. Can the probability of selecting two white balls from two bags ever be 100%?

No, the probability of selecting two white balls from two bags can never be 100%. This is because there is always a chance that at least one black ball will be selected from one of the bags, making the probability less than 100%.

## 4. What happens to the probability if one bag contains only white balls and the other bag contains only black balls?

If one bag contains only white balls and the other bag contains only black balls, the probability of selecting two white balls is 0%. This is because there are no white balls in one of the bags, making it impossible to select two white balls from two bags.

## 5. How does the probability change if the number of balls in each bag is increased?

The probability will change depending on the number of balls in each bag. If the number of balls in each bag is increased, the probability of selecting two white balls will decrease. This is because there are more balls in each bag, making it more likely to select a black ball. However, if the number of balls in each bag is increased, the probability of selecting two white balls will also increase. This is because there are more white balls in total, making it more likely to select two white balls.

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