Consider a population in which 80% of males and 60% of females are employed. In this population, 55% of individuals are females. If I pick five persons at random from this population, what is the probability that no more than one of those chosen is not employed?

The probability that no more than 1 person is not employed is 0.5077.

Step-by-step explanation:

Let's denote the events as:

E = a person is employed

M = a person is male

F = a person is a female.

Given:

P (E | M) = 0.80

P (E | F) = 0.60

P (F) = 0.55

P (M) = 1 - P (F) = 1 - 0.55 = 0.45

Compute the probability that a randomly selected person is employed as follows

P (E) = P (E | M) × P (M) + P (E | F) × P (F)

Then the probability that a randomly selected person is not employed is,

P (E') = 1 - P (E) = 1 - 0.69 = 0.31

Now let's assume X = number of employees that are not employed.

The sample selected is of size, n = 5.

Then the random variable

The probability function of Binomial distribution is:

Compute the probability that no more than 1 person is not employed as follows:

P (X ≤ 1) = P (X = 0) - P (X = 1)

Thus, the probability that no more than 1 person is not employed is 0.5077.

answer: a

step-by-step explanation: the answer is a due to the rule of symmetry; that a line of symmetry must divide a figure into 2 pieces that are the exact same, only reflected over a line of symmetry. b and c do not have lines of symmetry and would not divide the figure evenly, thus making the answer a.