Before 1918, wolves were shown as 55% male, 45% female. From 1918 to present they were shown as 70% male and 30% female. In a random sample of 12 wolves, what was the probability that 6 were male? What was the probibility that 6 or more were female? What was the probability fewer than 4 were female?

(b) Answer part A for period from 1918 to present

To calculate the probabilities, we need to use the binomial probability formula. The formula is as follows:

P(x) = C(n, x) * p^x * (1-p)^(n-x)

Where:
- P(x) is the probability of getting exactly x successes.
- C(n, x) is the number of combinations of n items taken x at a time.
- p is the probability of an individual item being a success.
- n is the total number of trials.

Now, let's calculate the probabilities for each part of the question:

(a) Before 1918:
- P(6 males out of 12 wolves):
Let's assume the probability of a wolf being male is p1 = 0.55.
Therefore, the probability of a wolf being female is 1 - p1 = 0.45.
Using the formula, the probability of getting exactly 6 males out of 12 wolves is:
P(6 males) = C(12, 6) * (0.55^6) * (0.45^6)

- P(6 or more females out of 12 wolves):
The probability of getting 6 or more females is equal to the sum of the probabilities of getting exactly 6, 7, 8, 9, 10, 11, or 12 females out of 12. We can calculate each probability individually and then sum them up.

- P(fewer than 4 females out of 12 wolves):
The probability of getting fewer than 4 females is equal to the sum of the probabilities of getting exactly 0, 1, 2, or 3 females out of 12. Again, we can calculate each probability individually and then sum them up.

(b) From 1918 to present:
Simply substitute the new probability values into the formulas mentioned in part (a) and calculate accordingly.

To solve these probabilities, we can use the binomial probability formula:

P(X = k) = nCk * p^k * q^(n-k)

where:
- P(X = k) is the probability of obtaining k successes
- nCk is the number of combinations of n items taken k at a time
- p is the probability of success
- q is the probability of failure
- n is the number of trials

First, let's calculate the probabilities for the period before 1918:

a) Probability of 6 male wolves in a random sample of 12:
n = 12 (number of trials)
k = 6 (number of successes)
p = 0.55 (probability of male)
q = 1 - p = 1 - 0.55 = 0.45 (probability of female)

P(X = 6) = 12C6 * 0.55^6 * 0.45^6

To calculate this, we can use the binomial coefficient formula:

12C6 = 12! / (6!*(12-6)!) = 924

Substituting the values into the formula:

P(X = 6) = 924 * 0.55^6 * 0.45^6

Calculating the result:

P(X = 6) ≈ 0.212

Therefore, the probability that 6 wolves are male is approximately 0.212.

b) Probability of 6 or more female wolves in a sample of 12:
We can find the probability of fewer than 6 female wolves and subtract it from 1.

P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)

We can calculate each term individually using the same formula as before.

P(X = 0) = 12C0 * 0.45^0 * 0.55^12
P(X = 1) = 12C1 * 0.45^1 * 0.55^11
P(X = 2) = 12C2 * 0.45^2 * 0.55^10
P(X = 3) = 12C3 * 0.45^3 * 0.55^9
P(X = 4) = 12C4 * 0.45^4 * 0.55^8
P(X = 5) = 12C5 * 0.45^5 * 0.55^7

Calculating these values and summing them:

P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)

P(X < 6) ≈ 0.218

Therefore, the probability that 6 or more wolves are female is approximately 0.218.

c) Probability of fewer than 4 female wolves in a sample of 12:
P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

Using the same formula as before, calculate each term individually:

P(X = 0) = 12C0 * 0.45^0 * 0.55^12
P(X = 1) = 12C1 * 0.45^1 * 0.55^11
P(X = 2) = 12C2 * 0.45^2 * 0.55^10
P(X = 3) = 12C3 * 0.45^3 * 0.55^9

Calculating these values and summing them:

P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

P(X < 4) ≈ 0.896

Therefore, the probability that fewer than 4 wolves are female is approximately 0.896.

Please note that these calculations are based on the assumption that the distribution of male and female wolves is independent and follows the given probabilities.

Prob(male) = .7

prob(female) = .3

a) prob(6 are male_
= C(12,6) (.7)^6 (.3)^6 = .07925

b) prob(6 or more are female)
= prob(6 F) + prob(7 F) + prob(8 F + .. prob(12 F_
=
C(12,6)(.3)^6 (.7)^6 + C(12,7) (.3)^7 (.7)^5 + ... + C(12,12) (.3)^12 (.7)^0
I will let you do the button pushing

prob(less than 4 are female)
= prob(3 F) + prob(2 F) + prob (1 F) + prob(0 F)
= C(12,3) (.3)^3 (.7)^9 + ... + C(12, 0) (.3)^0 (.7)^12
= .....