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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.

$\therefore p=0.1$ 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}.

$=1- P[X=0]$

$=1-$ $^4C_0$ $(0.1)^0(0.9)^4$

$=1-(0.9)^4$

Nice

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