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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.Â
​\( \therefore p=0.1 \)​ and ​\( q=1-p=1-0.1=0.9 \)​Â
Since, traveler makes 4 trips, n=4.Â
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}.Â
​\( =1- P[X=0] \)​
​\( =1- ^4C_0 (0.1)^0(0.9)^4 \)​
​\( =1-(0.9)^4 \)​
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