In a bowling league, the probability that the uniforms are all 1 color is 0,45, that 2 colors are used is 0.35, and that 3 or more colors are use is 0.20. if a sample of 12 uniforms is selected, find the probability that 5 contain only 1 color, 4 contain 2 colors, and 3 contain 3 or more colors.

We can use the multinomial probability formula to solve this problem. The multinomial probability formula is:

P(x1, x2, ..., xn) = n! / (x1! * x2! * ... * xn!) * p1^x1 * p2^x2 * ... * pn^xn

where:
n = total number of trials (in this case, 12 uniforms)
x1, x2, ..., xn are the number of trials resulting in each outcome (5, 4, and 3 uniforms)
p1, p2, ..., pn are the probabilities of each outcome (0.45, 0.35, and 0.20)

First, we need to calculate the factorials:

12! = 12 * 11 * 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 = 479001600
5! = 5 * 4 * 3 * 2 * 1 = 120
4! = 4 * 3 * 2 * 1 = 24
3! = 3 * 2 * 1 = 6

Now we can plug these values into the multinomial probability formula:

P(5, 4, 3) = (479001600) / (120 * 24 * 6) * 0.45^5 * 0.35^4 * 0.20^3
P(5, 4, 3) = (479001600) / (165888) * 0.45^5 * 0.35^4 * 0.20^3

Now we can calculate the powers:

0.45^5 ≈ 0.018
0.35^4 ≈ 0.015
0.20^3 ≈ 0.008

We can now multiply these values together:

P(5, 4, 3) = (479001600) / (165888) * 0.018 * 0.015 * 0.008
P(5, 4, 3) ≈ 2881.45 * 0.018 * 0.015 * 0.008
P(5, 4, 3) ≈ 0.006

Therefore, the probability that 5 uniforms contain only 1 color, 4 uniforms contain 2 colors, and 3 uniforms contain 3 or more colors is approximately 0.006.

To find the probability that 5 uniforms contain only 1 color, 4 uniforms contain 2 colors, and 3 uniforms contain 3 or more colors, we can multiply the individual probabilities together.

Let's calculate each probability:

Probability of 5 uniforms containing only 1 color:
This probability is given as 0.45.
Let's call this event A.

Probability of 4 uniforms containing 2 colors:
This probability is given as 0.35.
Let's call this event B.

Probability of 3 uniforms containing 3 or more colors:
This probability is given as 0.20.
Let's call this event C.

Now, we need to find the probability of event A happening 5 times, event B happening 4 times, and event C happening 3 times simultaneously.

Using the formula for the probability of multiple independent events occurring (assuming that there are no dependencies between the events), we can multiply the probabilities together.

P(A and B and C) = P(A) * P(A) * P(A) * P(A) * P(A) * P(B) * P(B) * P(B) * P(B) * P(C) * P(C) * P(C)

P(A and B and C) = 0.45^5 * 0.35^4 * 0.20^3

Calculating this expression yields the probability that 5 uniforms contain only 1 color, 4 contain 2 colors, and 3 contain 3 or more colors.

To find the probability that 5 uniforms contain only 1 color, 4 uniforms contain 2 colors, and 3 uniforms contain 3 or more colors out of a sample of 12 uniforms, we need to use the concept of permutations and combinations.

First, let's find the probability for each category:
1. Probability that 5 uniforms contain only 1 color:
We have 12 uniforms, and we need to choose 5 of them to contain only 1 color. The remaining uniforms will have different colors. So, we can use combinations to find the number of ways to choose 5 uniforms from 12. The formula for combinations is: nCr = n! / (r!(n-r)!), where n is the total number of items and r is the number of items being chosen.
In this case, n = 12 and r = 5. So, the number of ways to choose 5 uniforms out of 12 is 12C5 = 792. Now, we need to multiply this by the probability of uniforms having only 1 color, which is 0.45. Therefore, the probability that 5 uniforms contain only 1 color is 792 * 0.45.

2. Probability that 4 uniforms contain 2 colors:
Similarly, we need to choose 4 uniforms to contain 2 colors out of the remaining 7 uniforms (12 - 5 = 7). Again, we can use combinations to find the number of ways to choose 4 uniforms from 7. So, the number of ways to choose 4 uniforms is 7C4 = 35. We multiply this by the probability of uniforms having 2 colors, which is 0.35. Therefore, the probability that 4 uniforms contain 2 colors is 35 * 0.35.

3. Probability that 3 uniforms contain 3 or more colors:
We have chosen uniforms for 5 single-color uniforms and 4 two-color uniforms, which leaves us with 12 - 5 - 4 = 3 uniforms to choose from. Since all 3 of these uniforms contain 3 or more colors, the probability is given directly as 0.20.

Now, we multiply the probabilities for each category together because they are independent events:
Probability = (Number of ways for 5 single-color uniforms) * (Number of ways for 4 two-color uniforms) * (Probability for 3 multi-color uniforms)
Probability = (792 * 0.45) * (35 * 0.35) * 0.20

Now, you can substitute the values and calculate the final probability.