Question:
The probability that a ticketless traveler is caught during a trip is 0.1. If the traveler makes 4 trips , the probability that he/she will be caught during at least one of the trips is:
1. \( 1-(0.9)^4 \)
2. \( (1-0.9)^4 \)
3. \( 1-(1-0.9)^4 \)
4. \( (0.9)^4 \)
Answer:
The probability that a ticketless traveler is caught during a trip is 0.1.
\( \therefore p=0.1 \) and \( q=1-p=1-0.1=0.9 \)
Since, traveler makes 4 trips, n=4.
Let X be the random variable that the traveler is caught during a trip.
Hence, \( X \leadsto B(n=4, p=0.1) \) and \( X=0, 1, 2, 3, 4. \)
\( \therefore P[X=x]= ^4C_x (0.1)^x(0.9)^{4-x} \)
\( \therefore \) the probability that he/she will be caught during at least one of the trips
=1-{probability that he/she will be caught during at most one of the trips}.
\( =1- P[X=0] \)
\( =1- ^4C_0 (0.1)^0(0.9)^4 \)
\( =1-(0.9)^4 \)
