CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


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 (AB) = { 15, 30, 45, ...495}
We have:
nA=4983=166

nB=5005=100

nAB=49515=33      [∵ LCM of 3 and 5 is 15]

PA=166500, PB=100500 and PAB=33500

Now, required probability = P(a number is divisible by 3 or 5)
                                        = P (A ∪ B)
                                        = P(A) + P(B) - P(A ∩ B)
                                        = 166500+100500-33500=266-33500=233500

Mathematics
RD Sharma XI (2015)
Standard XI

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
Same exercise questions
View More


similar_icon
People also searched for
View More



footer-image