This is tricky--at first glance it appears to be a probability puzzle for mutually exclusive events since two people cannot occupy the same job (If events A and B are mutually exclusive, then the probability of A or B is simply:p(A or B) = p(A) + p(B) )

However, since the question asks simply for the probablitity that at least one of the seekers accepts the offer, then it seems correct to think of the job seekers as independent of each other. My guess, therefore, would be that there is a 60% chance that at least one of them accepts the job, since 60% of the job seekers accept on average. Mathematicians out there please chime in.

HI
I THINK it will be
= Probabilty candidate A accepts and Probability candidate B rejects + Probablity candidate b Accepts and Probablity candindate A rejects

=0.6*0.4+0.6*0.4
what do u think ?

jawad wrote:Hi

Please help with below probability question.

A company has the data which says that 40% of the individuals who were offered the job did not accept it. This year company offered the job to 2-people. What is the probability that atleast one of them will accept it.

Regrads ,Jawad

1-person_1_does_not_accept*person_2_does_not_accept

1-(40/100)(40/100) => 21/25

SEESAI wrote:HI
I THINK it will be
= Probabilty candidate A accepts and Probability candidate B rejects + Probablity candidate b Accepts and Probablity candindate A rejects

=0.6*0.4+0.6*0.4
what do u think ?

wrong. ignores the case that both accept.

My solution:

The first statement gives us the probability that a person refuses a job is 0.4.

When you see AT LEAST ONE type of probability questions, try converting it into NONE, and then subtract this probability value from 1 to get the desired answer.

So, probability that neither of those offered will accept the job is 0.4*0.4 (independent probabilty. hence A AND B conberts to P(A)*P(B)).

Therefore, probability that at least one will accept the offer is 1 - 0.4*0.4 = 0.84

What is the OA?