- #1
vcsharp2003
- 913
- 179
- 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.
(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:
