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Question

A natural number is chosen at random from amongst first 500. What is hte probability that the number so chosen is divisbile by 3 or 5 ?

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Solution

Since a number is chosen from first 500

n(S)=500

Let A be the event of choosing number divisble by 3= [3,6,9,...,498]

n(A)=166 a+(n1)d=4983+(n1)3=4983n=498n=166

P(A)=166500

B be the event of choosing number

[5,10,15,...,500]

n(B)= 100

[ a+(n-1)d=500]

P(B)=100500

[ 5+(n-1)5=500]

[5n=500n=100]

Also, (AB)={15,30,...,495}

n(AB)=33

⎢ ⎢ ⎢ ⎢a+(n1)d=49515+(n1)15=49515n=495n=3⎥ ⎥ ⎥ ⎥

P(AB)=33500

Now required probability = P(a number is divisible by 3 or 5) =P(AB)

P(AB)=P(A)+P(B)P(AB)

= 166500+10050033500=233500


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