Question

# 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?

Solution

## Let S be the sample space. Then n(S) = 500 ∴ Total number of elementary events = 500 Let A be the event where the number selected is divisible by 3 and B be the event where the number selected is divisible by 5. Then A = {3, 6, 9, 12, 15, ...498} and B = {5, 10, 15, 20, 25, ...500} and (A ∩ B) = { 15, 30, 45, ...495} We have: $n\left(A\right)=\frac{498}{3}=166$ $n\left(B\right)=\frac{500}{5}=100\phantom{\rule{0ex}{0ex}}$ $n\left(A\cap B\right)=\frac{495}{15}=33$      [∵ LCM of 3 and 5 is 15] Now, required probability = P(a number is divisible by 3 or 5)                                         = P (A ∪ B)                                         = P(A) + P(B) $-$ P(A ∩ B)                                         = $\frac{166}{500}+\frac{100}{500}-\frac{33}{500}=\frac{266-33}{500}=\frac{233}{500}$MathematicsRD Sharma XI (2015)Standard XI

Suggest Corrections

0

Similar questions
View More

Same exercise questions
View More

People also searched for
View More