Question

1. In a classroom on average only 3 students came to attend the class per day, find the probability for exactly 4 students to attend the classes tomorrow. 0.186 0.168 0.128 0.188 2. The probability that an eagle kills a rabbit in a day of hunting is 10%. Assume that results are independent for each day. What is the probability that the first successful hunt occurs more than 5 days? 0.5 0.6

Answer 1

To find the probability for exactly 4 students to attend the class tomorrow, we can use the binomial probability formula. The probability that the first successful hunt occurs more than 5 days can be calculated using the geometric probability formula.

Step-by-step explanation:

To find the probability for exactly 4 students to attend the class tomorrow, we can use the binomial probability formula. The formula is P(X=k) = C(n,k) * p^k * (1-p)^(n-k), where n is the number of trials, k is the number of successes, p is the probability of success on a single trial, and C(n,k) is the number of combinations of n items taken k at a time. In this case, n=1 (since it is a single day), k=4, p=0.03 (3 students out of 100), and C(n,k) = n!/(k!(n-k)!). Substituting these values into the formula:

P(X=4) = C(1,4) * 0.03^4 * (1-0.03)^(1-4)

Since C(1,4) = 1/4!, which is 1, the formula simplifies to:

P(X=4) = 0.03^4 * (1-0.03)^(-3)

Calculating this, we get P(X=4) = 0.000081.

Therefore, the probability for exactly 4 students to attend the class tomorrow is 0.000081

For the second question, the probability that the first successful hunt occurs more than 5 days can be calculated using the geometric probability formula. The formula is P(X>k) = (1-p)^k, where p is the probability of success and k is the number of trials until the first success occurs. In this case, p=0.10 (10% chance of a successful hunt), and k>5 (more than 5 days). Substituting these values into the formula:

P(X>5) = (1-0.10)^5

Calculating this, we get P(X>5) = 0.59049.

Therefore, the probability that the first successful hunt occurs more than 5 days is 0.59049

Answer 2

The probability of exactly 4 students attending tomorrow's class is 0, and the probability that the first successful hunt occurs more than 5 days is approximately 0.60149.

Step-by-step explanation:

To calculate the probability of exactly 4 students attending tomorrow's class, we need to use the binomial probability formula. The formula for the probability of exactly k successes in n independent trials is:

`P(X = k) = (n C k) * p^k * (1 - p)^(n - k)`

where n is the number of trials, k is the number of successes, p is the probability of success for each trial, and (n C k) represents the number of combinations of n items taken k at a time.

In this case, n = 3, k = 4, and p = 1/3 (since on average only 3 students attend per day).

Using the formula, we get:

`P(X = 4) = (3 C 4) * (1/3)^4 * (2/3)^(3 - 4)`

= 0 * (1/81) * (2/3) = 0

Therefore, the probability for exactly 4 students to attend the class tomorrow is 0.

For the second question, the probability that the first successful hunt occurs more than 5 days can be calculated using geometric probability. Since the hunt results are independent for each day, the probability of the first successful hunt occurring on any given day is 0.1.

The probability that the first successful hunt occurs more than 5 days is the complement of the probability that it occurs within the first 5 days. Therefore, the probability is:

`P(X > 5) = 1 - P(X < = 5) = 1 - [P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)]`

`P(X > 5) = 1 - [0.1 + (0.9 * 0.1) + (0.9^2 * 0.1) + (0.9^3 * 0.1) + (0.9^4 * 0.1)]`

P(X > 5) = 1 - [0.1 + 0.09 + 0.081 + 0.0729 + 0.06561]

P(X > 5) = 1 - 0.39851 = 0.60149

Therefore, the probability that the first successful hunt occurs more than 5 days is approximately 0.60149.