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

your answer would be 12

there are 25 people coming to the party

step-by-step explanation:

the text says that there will be 22 children at this party, 1 teacher, and 2 parents. so if you add them all together (22+1+2) it will equal 25. (=25)