1, 40% of the days tibs are available in the banchs cafeteria. Abdul is eating in banchs cafeteria for 10 days.
B, what is the probability that all 10 days tibs are available?
C,what is the probability that all 10 days no tibs are avaliable?
A, what is the probability that 3 of the 10 days tibs are avaliable?
2,suppose the probability that a house hold lives below the poverty line, is 30% in a certain area.an economist checks the poverty status of 20 house holds.A, what is the probability that 10 of the checked house holds live below the poverty line?
C, how many house holds, out of the 20,do you expect to live below the poverty line?
d, what is the variance of the expected number of house holds below the poverty line?
1. A. The probability of tibs being available on any given day is 0.4. Therefore, the probability of tibs being available for all 10 days is 0.4^10, which is approximately 0.0001.
B. This is the same probability as calculated in part A, which is approximately 0.0001.
C. The probability of no tibs being available on any given day is 0.6. Therefore, the probability of no tibs being available for all 10 days is 0.6^10, which is approximately 0.006.
1. A. The probability of a household living below the poverty line is 0.3. Therefore, the probability of 10 out of 20 households living below the poverty line can be calculated by using the binomial distribution formula. P(X = 10) = (20 choose 10)*(0.3)^10*(0.7)^10, which is approximately 0.018.
C. The expected number of households living below the poverty line can be calculated by multiplying the probability of a household living below the poverty line by the total number of households. E(X) = 0.3*20, which is equal to 6 households.
D. The variance of the expected number of households living below the poverty line can be calculated using the variance formula for a binomial distribution. Var(X) = npq, where n is the total number of trials (20), p is the probability of success on any given trial (0.3), and q is the probability of failure on any given trial (0.7). Var(X) = 20*0.3*0.7, which is equal to 4.2.
For question 1:
A) To find the probability that 3 out of the 10 days tibs are available, we need to use the binomial probability formula. The formula is:
P(X = k) = (nCk) * (p^k) * ((1-p)^(n-k))
where n is the number of trials (10 in this case), k is the number of successes (3 in this case), and p is the probability of success (40%, which in decimal form is 0.4).
Plugging in the values, we have:
P(X = 3) = (10C3) * (0.4^3) * (0.6^(10-3))
To calculate this, we need to find the combination value (nCk), which is calculated using:
nCk = n! / (k! * (n-k)!)
where ! denotes the factorial of a number.
Calculating the combination value:
10C3 = 10! / (3! * (10-3)!)
= 10! / (3! * 7!)
Note: n! means the factorial of n, which means multiplying all numbers from 1 to n.
Substituting the values:
10C3 = 10 * 9 * 8 / (3 * 2 * 1)
= 120
Now, substituting these values into the binomial probability formula:
P(X = 3) = 120 * (0.4^3) * (0.6^7)
Calculating this will give you the probability that 3 of the 10 days tibs are available.
B) To find the probability that all 10 days tibs are available, we raise the probability of tibs availability (0.4) to the power of 10:
P(all 10 days tibs available) = (0.4)^10
Calculating this gives you the probability that all 10 days tibs are available.
C) To find the probability that all 10 days no tibs are available, we use the complementary probability. The complementary probability is the probability of the event not happening.
P(all 10 days no tibs available) = 1 - P(all 10 days tibs available)
Calculating this gives you the probability that all 10 days no tibs are available.
For question 2:
A) To find the probability that 10 out of the 20 households live below the poverty line, we use the binomial probability formula. The formula is the same as in question 1.
The values we need are: n = 20 (number of trials), k = 10 (number of successes), and p = 0.3 (probability of success, which is living below the poverty line).
Similarly, calculate the combination value (20C10) and substitute all values into the binomial probability formula to find the probability.
C) To find the expected number of households living below the poverty line out of the 20, we multiply the total number of households (20) by the probability of living below the poverty line (0.3):
Expected number = Total number of households * Probability of living below poverty line
Calculating this gives you the expected number of households living below the poverty line.
D) To find the variance of the expected number of households below the poverty line, we use the formula:
Variance = np(1 - p)
where n is the number of trials (20) and p is the probability of success (0.3).
Substituting the values:
Variance = 20 * 0.3 * (1 - 0.3)
Calculating this will give you the variance of the expected number of households below the poverty line.