A bag contains 5 blue candies and 6 green candies. You randomly pick a piece of candy from the bag, observe the color, then put it back in the bag. You repeat this process 5 times. Find the following probabilities (rounded to the nearest thousandths):

a.) getting exactly
2 green candies. P(2 green)=

b.) getting exactly 1 blue candies. P(1 blue)=

a.) (6/11)^2 * (5/11)^3 = ?

b.) (5/11) * (6/11)^4 = ?

To find the probabilities, we need to use the concept of probability mass function (PMF).

a.) Getting exactly 2 green candies.
The total number of candies in the bag is 5 (blue) + 6 (green) = 11 candies.
The probability of picking a green candy on any one trial is 6/11, as there are 6 green candies out of the total 11 candies.
Since we are replacing the candy back into the bag after each trial, the probability of getting exactly 2 green candies is given by the binomial distribution formula:

P(2 green) = C(5,2) * (6/11)^2 * (5/11)^(5-2)

Using the combination formula C(n, r) = n! / (r!*(n-r)!), we can calculate:

P(2 green) = (5! / (2! * (5-2)!)) * (6/11)^2 * (5/11)^3

Calculating this, we find:

P(2 green) ≈ 0.290

Therefore, the probability of getting exactly 2 green candies (rounded to the nearest thousandths) is approximately 0.290.

b.) Getting exactly 1 blue candy.
Using the same logic as above, the probability of picking a blue candy on any one trial is 5/11, as there are 5 blue candies out of the total 11 candies.
The probability of getting exactly 1 blue candy is given by the binomial distribution formula:

P(1 blue) = C(5,1) * (5/11)^1 * (6/11)^(5-1)

Using the combination formula, we can calculate:

P(1 blue) = (5! / (1! * (5-1)!)) * (5/11)^1 * (6/11)^4

Calculating this, we find:

P(1 blue) ≈ 0.370

Therefore, the probability of getting exactly 1 blue candy (rounded to the nearest thousandths) is approximately 0.370.

To find the probability of getting a certain number of green and blue candies, we can use the concept of combinations.

a.) Getting exactly 2 green candies: P(2 green)
The probability of getting 2 green candies can be calculated as follows:

1. Find the total number of possible outcomes. In this case, you are repeating the process 5 times, and each time you have 2 choices (blue or green). So, the total number of outcomes is 2^5 = 32.

2. Find the number of favorable outcomes. Since we want exactly 2 green candies, we can choose 2 out of the 5 rounds to be green and the remaining 3 rounds to be blue. The number of ways to choose 2 out of 5 is given by the combination formula: C(5, 2) = 5! / (2! * (5-2)!) = 10.

3. Calculate the probability by dividing the number of favorable outcomes by the total number of possible outcomes: P(2 green) = 10 / 32 ≈ 0.312.

Therefore, the probability of getting exactly 2 green candies is approximately 0.312.

b.) Getting exactly 1 blue candy: P(1 blue)
The probability of getting 1 blue candy can be calculated in a similar way:

1. Find the total number of possible outcomes, which is still 2^5 = 32.

2. Find the number of favorable outcomes. In this case, we want exactly 1 blue candy, so we could choose any 1 out of the 5 rounds to be blue and the remaining 4 rounds to be green. The number of ways to choose 1 out of 5 is given by the combination formula: C(5, 1) = 5! / (1! * (5-1)!) = 5.

3. Calculate the probability by dividing the number of favorable outcomes by the total number of possible outcomes: P(1 blue) = 5 / 32 ≈ 0.156.

Therefore, the probability of getting exactly 1 blue candy is approximately 0.156.