Tickets numbered 1 to 20 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of 3 or 5?
A. $$\frac{{1}}{{2}}$$
B. $$\frac{{2}}{{5}}$$
C. $$\frac{{8}}{{15}}$$
D. $$\frac{{9}}{{20}}$$
Answer: Option D
Solution(By Examveda Team)
Here, S = {1, 2, 3, 4, ...., 19, 20}Let E = event of getting a multiple of 3 or 5
= {3, 6 , 9, 12, 15, 18, 5, 10, 20}
$$\therefore P\left( E \right) = \frac{{n\left( E \right)}}{{n\left( S \right)}} = \frac{9}{{20}}$$
This Question Belongs to Arithmetic Ability >> Probability
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Comments ( 10 )
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A box contains 4 white, 6 green, 2 red and 5 yellow pens. If 4 pens are picked at random, what is the probability that one of them is green, 2 are white and 1 is yellow?
Okay I got it
What is multiple of a number?
Hw 9 comes
How
This is a Joint Probability question and formula to calculate joint probability is :
P(A or B) = P(A) + P(B) - P(A and B)
Here P(A) = Value of Multiple of 3 are {3,6,9,12,15,18} / Sample space = 6/20
P(B) = Value of Multiple of 5 are {5,10,15,20} / Sample space = 4/20
P(A and B) = Common favorable events between A and B are {15} = 1/20
As per formula P(A or B) = 6/20 + 4/20 - 1/20 = 9/20
Where is 15? multiple of 5
5 will have to be there, and then the answer would be 2/5.
Why don't you guys just correct it.
3 is not a multiple of 3. so why 3 is here. And if it is why 5 is not?
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