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# 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:

## 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. â€‹$$1-(0.9)^4$$â€‹
2. â€‹$$(1-0.9)^4$$â€‹
3. â€‹$$1-(1-0.9)^4$$â€‹
4. â€‹$$(0.9)^4$$â€‹

The probability that a ticketless traveler is caught during a trip is 0.1.Â
â€‹$$\therefore p=0.1$$â€‹ and â€‹$$q=1-p=1-0.1=0.9$$â€‹Â
Since, traveler makes 4 trips, n=4.Â
Let X be the random variable that the traveler is caught during a trip.
Hence, â€‹$$X \leadsto B(n=4, p=0.1)$$â€‹ and â€‹$$X=0, 1, 2, 3, 4.$$â€‹
â€‹$$\therefore P[X=x]= ^4C_x (0.1)^x(0.9)^{4-x}$$â€‹
â€‹$$\therefore$$â€‹Â the probability that he/she will be caught during at least one of the trips
=1-{probability thatÂ he/she will be caught during at most one of the trips}.Â
â€‹$$=1- P[X=0]$$â€‹
â€‹$$=1-Â ^4C_0Â (0.1)^0(0.9)^4$$â€‹
â€‹$$=1-(0.9)^4$$â€‹
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