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:

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. $1-(0.9)^4$  
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}. 
$=1- P[X=0]$ 
$=1-$ $^4C_0$ $(0.1)^0(0.9)^4$ 
$=1-(0.9)^4$ 

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