Question

A candidate is selected for interview for 3 posts. For the first post, there are five candidates, for the second there are 8 and for the third there is 7. What is the chance for his getting at least one post?

a) ⅕
b) ⅗
c) ⅖
d) ⅘​

Answer 1

Answer:

Let A, B and C denote the events that the candidate gets the first , second and third post respectively.ᴛ

$\sf \therefore \: P(A)= \frac{1}{3}$

$\sf \longrightarrow \: P(A)=1−P(A)= \frac{2}{3}$

$\sf \: P(B)= \frac{1}{4}$

$\sf \implies \: P(C)= \frac{1}{2}$

$\sf \: P(C)= \frac{1}{2}$

Now , probability that candidate will get at least one post is P(A∪B∪C)

$\sf 1−P(A∪B∪C)$

$\sf1−P(A∩B∩C)$

$\sf1−P(A)P(B)P(C)$

Since events A, B, C are independent ⇒ A , B , C , are also independent)

$= \tt1 - \times \frac{2}{3} \times \frac{3 }{4} \times \frac{1}{2} = \frac{3}{4}$

Answer 2

$\huge \underline \mathtt \pink{❥ᴀnsᴡᴇʀ}$

Let A, B and C denote the events that the candidate gets the first , second and third post respectively.ᴛ

$\sf \therefore \: P(A)= \frac{1}{3}∴P(A)=31</p><p></p><p>\sf \longrightarrow \: P(A)=1−P(A)= \frac{2}{3}⟶P(A)=1−P(A)=32</p><p></p><p>\sf \: P(B)= \frac{1}{4}P(B)=41</p><p></p><p>\sf \implies \: P(C)= \frac{1}{2}⟹P(C)=21</p><p></p><p>$

Now , probability that candidate will get at least one post is P(A∪B∪C)

$\sf 1−P(A∪B∪C)1−P(A∪B∪C)</p><p></p><p>\sf1−P(A∩B∩C)1−P(A∩B∩C)</p><p></p><p>\sf1−P(A)P(B)P(C)  1−P(A)P(B)P(C)  </p><p>$

Since events A, B, C are independent ⇒ A , B , C , are also independent)

$\tt1 - \times \frac{2}{3} \times \frac{3 }{4} \times \frac{1}{2} = \frac{3}{4}=1−×32×43×21=43</p><p></p><p>$