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 answer

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.