Aptitude - Probability - Discussion

From 1 to 20, numbers which are divisible by 3 are 3, 6, 9, 12, 15 & 18 -> total 6.
Numbers divisible by 5 are 5, 10, 15, 20 total 4.

So, 6 + 4 = 10. But 15 is counted twice. Therefore 10 - 1= 9. Result is 9/20.

P(A) + P(B) - P(A and B).
6 + 4 - 1 = 9.
P(E) = n(E)/n(S).
P(E) = 9/20.

From 1 to 340, the probability of getting multiples of 8 and 5 is? Please answer this question.

Find the probability that a number from 1 to 300 is divisible by 3 or 7.

Can anyone answer this question.

According to me, From 1 to 20, numbers its 1, 2, 3, 4....., 19, 20 then number which is a multiple of 3 or 5.

Lets take 3 (3, 6, 9, 12, 15, 18 and 5 = (5, 10, 15, 20).

So in 3 we have 6 and 5 we have 4 so it's 6/20 and 4/20 and 1/20.

Solution = 6 + 4 - 1/20 == 9/20 answer.

According to me, From 1 to 20, numbers its 1, 2, 3, 4....., 19, 20 then number which is a multiple of 3 or 5.

Lets take 3 (3, 6, 9, 12, 15, 18 and 5 = (5, 10, 15, 20).

So in 3 we have 6 and 5 we have 4 so it's 6/20 and 4/20 and 1/20.

Solution = 6 + 4 - 1/20 == 9/20 answer.

Ans is 9/20 because,

P(A) + P(B) - P(A andB) = P(A OR B )

For p (a) outcome are divide to 20 for getting multiple of 3 we get in rational left after point so 6.
Then also of 5 we get 4 numbers can multiple of five.
After it puts in probability case such as 6/20 + 4/20 - 1/20.
1/20 comes to get no. of both that is only one number 15 and also use short cut as take the LCM of 3&5 we get 15, divide 20 by it to get some thing 1. Left fraction we get 1/ 20 or 1 out of 20 put all in formula.

6/20 + 4/20 - 1/20 = 9/20 answer.

Please help me to solve this-

A number is selected at random from first 50 number. Find the probability that it is a multiple of 3 and 4.

No, here 15 is repeated in 3&5 so the answer is 2/5.

How can we calculate the same problem with big numbers? i.e if tickets numbered 1-200.