A bag contains 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 ?
Try the rest now.
10 balls at start
1 prob black, white, red
= 3/10 * 4/9 * 5/8
2 prob black, red, white
= 3/10 * 5/9 * 4/8 same as above
3 prob white, black, red
= 4/10 * 3/9 * 5/8 same again
4 prob white, red , black
same
5 prob red , white, black
same
6 prob red, black, white
same
so
6 * 3/10 * 4/9 * 5/8
a) Well, let's calculate the probability of obtaining one ball of each color. There are three scenarios where this can happen: black, white, and red in any order.
The probability of drawing a black ball first is 3/12 (since there are 3 black balls out of 12 total). After removing the black ball, the probability of drawing a white ball is 4/11 (since there are 4 white balls remaining out of the remaining 11 balls). Finally, the probability of drawing a red ball is 5/10 (since there are 5 red balls remaining out of the remaining 10 balls).
To find the overall probability, we multiply the individual probabilities together:
(3/12) * (4/11) * (5/10) = 60/1320 = 1/22
So, the probability of obtaining one of each color is 1/22.
b) Now let's calculate the probability of getting at least two red balls. There are two scenarios where this can happen: 2 red balls and 3 red balls.
The probability of drawing 2 red balls can be calculated as follows:
- Drawing a red ball first, which has a probability of 5/12
- Then drawing another red ball, which now has a probability of 4/11 (since there is one less red ball remaining)
Multiply these together to get:
(5/12) * (4/11) = 20/132
The probability of drawing 3 red balls can be calculated as follows:
- Drawing a red ball first, which has a probability of 5/12
- Then drawing another red ball, which now has a probability of 4/11 (since there is one less red ball remaining)
- Finally, drawing a third red ball, which has a probability of 3/10 (since there are now two less red balls remaining)
Multiply these probabilities together to get:
(5/12) * (4/11) * (3/10) = 60/1320 = 1/22
To find the overall probability of getting at least two red balls, we add the probabilities of these two scenarios together:
20/132 + 1/22 = 22/132 = 1/6
So, the probability of getting at least two red balls is 1/6.
I hope that answers your question! Just remember, if these probabilities start causing you stress, I'm always here to lighten the mood with a joke.
To find the probability of obtaining a specific outcome, we first need to determine the total number of possible outcomes.
a) Probability of obtaining one of each color:
To find the probability of this outcome, we need to calculate the number of ways we can choose one black ball, one white ball, and one red ball, and then divide it by the total number of possible outcomes.
1. Calculate the number of ways to choose one ball of each color:
Let's start with the black ball. There are 3 black balls in total, so we can choose one black ball in 3 different ways.
For the white ball, we have 4 white balls, giving us 4 different choices.
Lastly, there are 5 red balls, so we have 5 different options for choosing one red ball.
2. Multiply the number of choices for each color:
To find the total number of ways to choose one ball of each color, we multiply the number of choices for each color together: 3 black balls × 4 white balls × 5 red balls = 60.
3. Calculate the total number of possible outcomes:
When three balls are removed without replacement, the total number of possible outcomes is the number of ways to choose any three balls from the total number of balls available. In this case, there are a total of 12 balls available initially. So the total number of possible outcomes is given by the number of ways to choose 3 balls out of 12, denoted as C(12,3) or 12 choose 3.
C(12,3) = 12! / (3! × (12 - 3)!) = 12! / (3! × 9!) = (12 × 11 × 10) / (3 × 2) = 220.
4. Calculate the probability:
Finally, we divide the number of favorable outcomes (when we obtain one of each color) by the total number of possible outcomes, and that gives us the probability.
Probability of obtaining one of each color = Number of favorable outcomes / Total number of possible outcomes
= 60 / 220
= 3 / 11 (approximately 0.2727, rounded to four decimal places)
Therefore, the probability of obtaining one black ball, one white ball, and one red ball is approximately 0.2727 or 27.27%.
b) Probability of obtaining at least two red balls:
To find the probability of this outcome, we need to calculate the number of ways we can choose two red balls (plus one ball of any other color) and the number of ways we can choose three red balls, and then divide it by the total number of possible outcomes.
1. Calculate the number of ways to choose two red balls (plus one ball of any other color):
There are 5 red balls in total, so we can choose two red balls in C(5,2) = 5! / (2! × (5 - 2)!).
C(5,2) = 5! / (2! × (5 - 2)!) = 5! / (2! × 3!) = (5 × 4) / (2 × 1) = 10.
For the remaining ball, it can be of any color except red. So, we have a total of 12 - 5 = 7 balls to choose from.
2. Calculate the number of ways to choose three red balls:
Since there are only 5 red balls available, it is not possible to choose three red balls without replacement.
3. Calculate the total number of possible outcomes (as calculated in part a):
Total number of possible outcomes = C(12,3) = 220.
4. Calculate the probability:
Now, we add the number of favorable outcomes for obtaining at least two red balls (10) to the number of favorable outcomes for obtaining three red balls (0). Then we divide the sum by the total number of possible outcomes.
Probability of obtaining at least two red balls = (Number of favorable outcomes with two red balls + Number of favorable outcomes with three red balls) / Total number of possible outcomes
= (10 + 0) / 220
= 10 / 220
= 1 / 22 (approximately 0.0455, rounded to four decimal places)
Therefore, the probability of obtaining at least two red balls is approximately 0.0455 or 4.55%.
A comittee consist of 6 men and 4 women. A subcomittee is made by randomly choosing three of the comittee members. What is the probability that
a they are all women
b two of them are men?