Question

20% of the applicants for a certain sales position are fluent in both Chinese and Spanish. Suppose that four jobs requiring fluency in Chinese and Spanish are open. Find the probability that two unqualified applicants are interviewed before finding the fourth qualified applicant, if the applicants are interviewed sequentially and at random.

Answer 1

To find the probability that two unqualified applicants are interviewed before finding the fourth qualified applicant, we can use the concept of geometric probability. The probability is 0.1024.

Step-by-step explanation:

To find the probability that two unqualified applicants are interviewed before finding the fourth qualified applicant, we can use the concept of geometric probability. Geometric probability is used when we want to find the probability of a certain event occurring for the first time after a certain number of independent trials.

In this case, there are 4 jobs requiring fluency in Chinese and Spanish and 20% of the applicants are fluent in both languages. Therefore, the probability of an applicant being qualified is 0.2. The probability of an applicant being unqualified is 1 - 0.2 = 0.8.

Now, let's calculate the probability:

1. The probability that the first applicant is unqualified is 0.8.
2. Given that the first applicant is unqualified, the probability that the second applicant is unqualified is also 0.8.
3. Given that the first two applicants are unqualified, the probability that the third applicant is unqualified is again 0.8.
4. Given that the first three applicants are unqualified, the probability that the fourth applicant is qualified is 0.2.

Therefore, the probability that two unqualified applicants are interviewed before finding the fourth qualified applicant is:

P = (0.8)(0.8)(0.8)(0.2) = 0.1024

Answer 2

The probability that two unqualified applicants are interviewed before finding the fourth qualified applicant = 0.164

Step-by-step explanation:

Let,

The number of unqualified applicants are interviewed before finding the fourth qualified applicant = x

So,

X be the negative Binomial (n = 4 , P =
`(20)/(100) = 0.2` )

As , we know that

P(X = n ) = ⁿ⁺⁴⁻¹Cₙ × (1 - 0.2 )⁴ × (0.2)ⁿ for n = 0, 1, 2, ....

As we have to find the probability that two unqualified applicants are interviewed before finding the fourth qualified applicant

⇒n = 2

∴

P(X = 2 ) = ²⁺⁴⁻¹C₂ × (1 - 0.2 )⁴ × (0.2)² for n = 0, 1, 2, ....

= ⁵C₂ × (0.8 )⁴ × (0.2)²

=
`(5!)/(2! (5-2)!)` × 0.41 × 0.04

=
`(5.4.3!)/(2.1! (3)!)` × 0.0164

=
`(5.4)/(2)` × 0.0164

= 10 × 0.0164 = 0.164

∴ we get

The probability that two unqualified applicants are interviewed before finding the fourth qualified applicant = 0.164