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Sagot :
Answer:
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 = [tex]\frac{20}{100} = 0.2[/tex] )
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
= [tex]\frac{5!}{2! (5-2)!}[/tex] × 0.41 × 0.04
= [tex]\frac{5.4.3!}{2.1! (3)!}[/tex] × 0.0164
= [tex]\frac{5.4}{2}[/tex] × 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
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