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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:

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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3 COMMENTS

  1. This is rightly named as the age of traveler-centricity and with the evolution of the new era of personalized travel; it is leading to research and development of a host of new so-called intelligent services. The command-and-control perspectives of traveling have changed a lot from the past and the focus has shifted more on the traveler and the productivity of each trip. It has become essential to maintain that the travelers have the greatest return on investment on each trip. Travel 101

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