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

Asked by Aaryan | 1 year ago |  148

##### Solution :-

Given: Sample space is the set of first 500 natural numbers.

n (S) = 500

Let ‘A’ be the event of choosing the number such that it is divisible by 3

n (A) = [$$\dfrac{ 500}{3}$$]

= [166.67]

= 166 {where [.] represents Greatest integer function}

P (A) =$$\dfrac{ n (A) }{ n (S)}$$

$$\dfrac{166}{500}$$

$$\dfrac{83}{250}$$

Let ‘B’ be the event of choosing the number such that it is divisible by 5

n (B) = [$$\dfrac{500}{5}$$]

= 

= 100 {where [.] represents Greatest integer function}

P (B) = $$\dfrac{ n (B) }{ n (S)}$$

= $$\dfrac{ 100}{500}$$

$$\dfrac{ 1}{5}$$

Now, we need to find the P (such that number chosen is divisible by 3 or 5)

P (A or B) = P (A ∪ B)

By using the definition of P (E or F) under axiomatic approach (also called addition theorem) we know that:

P (E ∪ F) = P (E) + P (F) – P (E ∩ F)

P (A ∪ B) = P (A) + P (B) – P (A ∩ B)

[Since, we don’t have value of P(A ∩ B) which represents event of choosing a number such that it is divisible by both 3 and 5 or we can say that it is divisible by 15.]

n(A ∩ B) = [$$\dfrac{ 500}{15}$$]

= [33.34]

= 33

P (A ∩ B) = $$\dfrac{ n(A ∩ B) }{n (S)}$$

=$$\dfrac{ 33}{500}$$

P (A ∪ B) = P (A) + P (B) – P (A ∩ B)

= $$\dfrac{ 83}{250} + \dfrac{1}{5} –\dfrac{ 33}{500}$$

=$$\dfrac{ [166 + 100 – 33]}{500}$$

$$\dfrac{ 233}{500}$$

Answered by Aaryan | 1 year ago

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