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]

= 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 agoOne number is chosen from numbers 1 to 100. Find the probability that it is divisible by 4 or 6?

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