Question

81000 Persons are visiting a football stadium. 2/3 of them are getting controlled at the entry. Averagely 100 persons are trying to smuggle pyrotechnics. But it also could be only 50.
1. You are trying to smuggle pyrotechnics into the stadium. What is the probability that you are not getting controlled at the entry?
2. You are trying to smuggle pyrotechnics into the stadium. What is the probability that you are successfull?

Answer 1

Totla people = 8100

P(C)=2/3

P(NC)=1-2/3=1/3

P(S)=100/81000

P(S|C)=50/8100

P(S|NC)=50/8100 (As 50 people were successful in succeeding)

1)Now we have to use Bay's theorm to find the required probability

P(C|S)= P(S|C) * P(C)/P(S)

= (50/8100*2/3)/(100/8100) = 1/3 =0.33333

So 1/3 is the probability that you are not getting controlled at the entry evenif you are trying to smuggle pyrotechnics into the stadium

2) Now You are trying to smuggle pyrotechnics into the stadium and we need to find the probability that you are successfull

P(S|NC) = P(NC|S)*P(S)/P(NC)

= (50/8100*100/8100)/(1/3) =0.0002286237

So 0.0002286237 is the probability that you are successfull given that You are trying to smuggle pyrotechnics into the stadium

Hope the above answer has helped you in understanding the problem. Please upvote the ans if it has really helped you. Good Luck!!