Question

In company XYZ, 30 percent of the workers take public transportation daily to go to work.

1. In a sample of 10 workers, what is the probability that 3 workers take public transportation to work daily?

a. .900
b. .267
c. .100
d. 1.20

2.In a sample of 10 workers, what is the probability that at least 3 workers take public transportation to work daily?

a. .767
b. .383
c. .617
d. .100

3.In a sample of 10 workers, what is the probability that, at most, 2 workers take public transportation to work daily?

a. .233
b. .121
c. .383
d. .149

Answer 1

Answer on Question #52392 – Math – Statistics and Probability

In company XYZ, 30 percent of the workers take public transportation daily to go to work.

$$p = 0.3.$$ $$P(X = k) = \frac{n!}{k! (n - k)!} p^k (1 - p)^{n - k}$$

1. In a sample of 10 workers, what is the probability that 3 workers take public transportation to work daily?

a. .900

b. .267

c. .100

d. 1.20

Solution

Choose the binomial distribution with parameters $(n,p) = (10,0.3)$ and

$$P(X = 3) = \frac{10!}{3! (10 - 3)!} 0.3^3 (1 - 0.3)^{10 - 3} = 0.267.$$

Answer: b. .267.

2. In a sample of 10 workers, what is the probability that at least 3 workers take public transportation to work daily?

a. .767

b. .383

c. .617

d. .100

Solution

Choose the binomial distribution with parameters $(n,p) = (10,0.3)$ and probability of complement is calculated by

$$P(X \geq 3) = 1 - P(X < 3).$$

By addition rule for probability of mutually exclusive events,

$$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2),$$

where

$$P(X = 0) = \frac{10!}{0! (10 - 0)!} 0.3^0 (1 - 0.3)^{10 - 0} = 0.0282.$$ $$P(X = 1) = \frac{10!}{1!(10 - 1)!} 0.3^1 (1 - 0.3)^{10 - 1} = 0.1211$$ $$P(X = 2) = \frac{10!}{2!(10 - 2)!} 0.3^2 (1 - 0.3)^{10 - 2} = 0.2335.$$

Therefore,

$$P(X \geq 3) = 1 - 0.0282 - 0.1211 - 0.2335 = 0.617.$$

Answer: c. .617.

3. In a sample of 10 workers, what is the probability that, at most, 2 workers take public transportation to work daily?

a. .233

b. .121

c. .383

d. .149

Solution

Probability of complement is given by

$$P(X \leq 2) = 1 - P(X \geq 3) = 1 - 0.617 = 0.383.$$

Answer: c. .383.

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