Can someone help me answer this question, so I would know what to do for a question like this next time?
Thank you!
A box contains 10 red chips, 20 blue chips, and 30 green chips. If 5 chips are drawn from the box, find the probability of drawing
a.) exactly 4 green
b.) at least 2 red
c.) at most 2 blue
I will assume the chips are not returned after being selected.
a)
You want the situation
GGGGN and its other 4 variations, where N stands for non-green
Prob(GGGGN) = (30/60)(29/59)(28/58)(27/57)(30/56)
= 135/4484
But there are C(5,4) of these
Prob(exactly 4 green) = 5(135/4484) = 675/4484
b) at least 2 red ---> exlude 0 red and 1 red
prob(zero red) = prob(NNNNN)
= (50/60)(49/59)(48/58)(47/57)(46/56)
= (50x49x48x47x46)/(60x59x58x57x56)
= 117/4484
prob(1 red) = 5x(50/60)(49/59)(48/58)(47/57)(10/56)
= .......
prob(at least 2 red)
= 1 - (prob(zero red) + prob(1 red) )
= ...
c) at most 2 blue
= prob(zero blue) + prob(1 blue) + prob(2 blue)
check my arithmetic and typing of the set-up.
I am only on my second cup of coffee.
Of course! I'll be happy to help you answer this question and explain how to solve similar problems in the future.
In order to find the probabilities, we need to understand the concept of probability. Probability is a measure of how likely an event is to occur. It is computed by dividing the number of favorable outcomes by the total number of possible outcomes.
Let's break down the problem and find the probabilities step by step.
a.) To find the probability of drawing exactly 4 green chips, we need to calculate the number of favorable outcomes and divide it by the total number of possible outcomes.
The number of ways to draw 4 green chips out of 30 green chips is denoted as "nCr," which represents the number of combinations. In this case, we have 30 green chips and we want to choose 4, so the calculation would be:
30C4 = 27,405
The total number of ways to draw 5 chips from the box, regardless of color, is calculated as:
60C5 = 142,506
Therefore, the probability of drawing exactly 4 green chips can be found by dividing the number of favorable outcomes by the total number of outcomes:
Probability = 27,405 / 142,506 ≈ 0.192
So, the probability of drawing exactly 4 green chips is approximately 0.192.
b.) To find the probability of drawing at least 2 red chips, we need to consider two scenarios: drawing exactly 2 red chips and drawing more than 2 red chips. We'll calculate these probabilities separately and then add them together.
The probability of drawing exactly 2 red chips can be calculated in a similar way as we did in part a. The number of favorable outcomes is given by:
10C2 = 45
The total number of outcomes is still 142,506.
The probability of drawing more than 2 red chips can be found by calculating the probability of drawing 3, 4, or 5 red chips. We add these probabilities together to get the total probability.
10C3 + 10C4 + 10C5 = 120 + 210 + 252 = 582
So, the probability of drawing at least 2 red chips can be found by adding the probabilities of the two scenarios:
Probability = (45 + 582) / 142,506 ≈ 0.004
Therefore, the probability of drawing at least 2 red chips is approximately 0.004.
c.) To find the probability of drawing at most 2 blue chips, we need to calculate the probabilities of drawing 0, 1, or 2 blue chips and add them together.
The probability of drawing 0 blue chips can be calculated as:
(30C5 + 10C5) / 142,506 ≈ 0.198
The probability of drawing 1 blue chip can be calculated as:
(30C4 * 20C1 + 30C3 * 20C2) / 142,506 ≈ 0.468
The probability of drawing 2 blue chips can be calculated as:
(30C3 * 20C2) / 142,506 ≈ 0.256
To find the probability of drawing at most 2 blue chips, we add these probabilities together:
Probability = 0.198 + 0.468 + 0.256 ≈ 0.922
Therefore, the probability of drawing at most 2 blue chips is approximately 0.922.
I hope this helps! Let me know if you have any further questions.