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  • The probability that a randomly selected number from the set  <img alt="\{1,2,3, \ldots, 100\}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5C%7B1%2C2%2C3%2C%20%5Cldots%2C%20100

The probability that a randomly selected number from the set  \{1,2,3, \ldots, 100\} will be divisible by 3 is  33.33 \%  What is the probability that it will be divisible by 5 ?

Option: 1

0.40


Option: 2

0.53


Option: 3

0.62


Option: 4

0.71


Answers (1)

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Solution: The probability that a randomly selected number from the set \{1,2,3, \ldots, 100\}  will be divisible by 3 is 33.33 \% , which means that there are 33 numbers divisible by 3 in the set.
The probability that a randomly selected number from the set will be divisible by 5 is 20 %, which means that there are 20 numbers divisible by 5 in the set.

Since the sets are not mutually exclusive, we cannot simply add the two probabilities to get the probability that a randomly selected number will be divisible by 3 or 5 . Instead, we need to use the following formula:
\mathrm{P(A \text { or } B)=P(A)+P(B)-P(A \text { and } B) }

Where,

-\mathrm{A}  is the event that the number is divisible by 3
- \mathrm{B}  is the event that the number is divisible by 5
- \mathrm{P(A)}  is the probability of event \mathrm{A}
- \mathrm{P(B)}  is the probability of event \mathrm{B}
- \mathrm{P(A and B)}is the probability of events \mathrm{A} and \mathrm{B} occurring simultaneously
In this case, \mathrm{P}(\mathrm{A}) and \mathrm{P}(\mathrm{B)} is zero because no number can be divisible by both 3 and 5 at the same time. So the formula becomes:
\mathrm{P(A \text { or } B)=0.3333+0.20-0=0.5333 }
Therefore, the probability that a randomly selected number from the set will be divisible by 3 or 5 is 53.33 \%

Posted by

Devendra Khairwa

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