A bag contain three black balls,four white balls and five red balls.Three balls are removed without replacement.What is the probability of obtaining
a)one of each colour
b)at least two red balls?
So you want BWR
that particular order's prob = (3/12)(4/11)(5/10) = 1/22
but the BWR can be arranged in 3! or 6 ways.
so Prob(your event) = 6/22 = 3/11
b) could RRB or RRW
prob(RRB in that order) = (5/12)(4/11)(3/10) = 1/22
we it could have been RRB, RBR, BRR, so
prob(of RRB in any order) = 3/22
do the same for RRW, then add them up
To calculate the probability of obtaining one of each color when three balls are removed without replacement, we need to find the favorable outcomes (the number of ways we can choose one black, one white, and one red ball) and divide it by the total number of possible outcomes (the number of ways we can choose any three balls from the bag).
a) Probability of obtaining one of each color:
1. Calculate the number of ways to choose one black, one white, and one red ball:
- Number of ways to choose one black ball: 3 (because there are three black balls in the bag).
- Number of ways to choose one white ball: 4 (because there are four white balls in the bag).
- Number of ways to choose one red ball: 5 (because there are five red balls in the bag).
2. Multiply these numbers together: 3 × 4 × 5 = 60.
3. Calculate the total number of ways to choose any three balls from the bag:
- Total number of balls in the bag = 3 + 4 + 5 = 12.
- Number of ways to choose any three balls from 12 = 12C3 = 220 (using combinations formula).
4. Divide the number of favorable outcomes by the total number of possible outcomes:
- Probability = 60/220 = 3/11 (approximately 0.273).
Therefore, the probability of obtaining one of each color is 3/11 (or approximately 0.273).
b) To calculate the probability of obtaining at least two red balls when three balls are removed without replacement, we need to find the favorable outcomes (the number of ways we can choose two or three red balls) and divide it by the total number of possible outcomes.
1. Calculate the number of ways to choose two red balls:
- Number of ways to choose two red balls: 5C2 = (5!)/[(2!)(5-2)!] = 10 (using combinations formula).
2. Calculate the number of ways to choose three red balls:
- Number of ways to choose three red balls: 5C3 = (5!)/[(3!)(5-3)!] = 10 (using combinations formula).
3. Add these two numbers together: 10 + 10 = 20.
4. Calculate the total number of possible outcomes (choosing any three balls from the bag): 12C3 = 220 (using combinations formula).
5. Divide the number of favorable outcomes by the total number of possible outcomes:
- Probability = 20/220 = 1/11 (approximately 0.091).
Therefore, the probability of obtaining at least two red balls is 1/11 (approximately 0.091).