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.

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 = )

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  

               = × 0.41 × 0.04

               = × 0.0164

               = × 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

2.56 is the width of the garden

step-by-step explanation:

true !

step-by-step explanation: