Question

While going from Mumbai to Pune by car, the probability that a person will be stopped by each of the 3 policemen who control the traffic at three different places is 0.6. What is the probability that
a) all the 3 policemen will stop the person?
b) none of the policemen will stop the person?
c) the first policeman will stop the person but not the other two?
d) at least two policemen will stop the person?
e) exactly two policemen will stop the person?
What assumptions do you make in finding the above probabilities?

Answer 1

Solution:

At each place Probability of stopping by Police man is
$p(E1) = p(E2) = p(E3) = 0.6$
So,probability of not stopping by Police man at each place

$p(\bar{E1}) = p(\bar E2) = p(\bar E3) = 0.4$
a) all the 3 policemen will stop the person?

Probability of all 3 traffic police officer stop that person:
$p(a) = p(E1) \times p(E2) \times p(E3) \\ \\ = 0.6 \times 0.6 \times 0.6 \\ \\ = \frac{27}{125} \\ \\ p(a)= 0.216$
b) none of the policemen will stop the person?

$p(b) = p(\bar E1) \times p(\bar E2) \times p(\bar E3) \\ \\ = 0.4 \times 0.4 \times 0.4 \\ \\ = \frac{8}{125} \\ \\ = 0.064$
c) the first policeman will stop the person but not the other two?

$p(c) = p(E1)p(\bar E2)p(\bar E3) \\ \\ = 0.6 \times 0.4 \times 0.4 \\ \\ = 0.096$
d) at least two policemen will stop the person?

$p(d) = p(E1) p(E2) p(\bar E3) + p(E1) p(E2) p(E3) + p(\bar E1) p(E2) p( e3) + p(E1) p(\bar E2) p(E3) \\ \\ = 0.6 \times 0.6 \times 0.4 + 0.6 \times 0.6 \times 0.6 + 0.4 \times 0.6 \times 0.6 + 0.6 \times 0.4 \times 0.6 \\ \\ = 0.144 +0.216 + 0.144 + 0.144 \\ \\ = 0.648$

e) exactly two policemen will stop the person?

$p(e) = p(E1) p(E2) p(\bar E3) + p(E1) p(\bar E2) p( E3) + p(\bar E1) p(E2) p(E3) \\ \\ = 0.6 \times 0.6 \times 0.4 + 0.6 \times 0.4 \times 0.6 + 0.4 \times 0.6 \times 0.6 \\ \\ = 0.144 + 0.144 + 0.144 \\ \\ = 0.432$
Assumption I had made that Probability of stopping by Police is p(E1),p(E2) and p(E3) at three different places respectively.