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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?
Answer & Solution
Correct Answer:
Option
D
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}
$$\therefore P\left( E \right) = \frac{{n\left( E \right)}}{{n\left( S \right)}} = \frac{9}{{20}}$$
Let E = event of getting a multiple of 3 or 5
= {3, 6 , 9, 12, 15, 18, 5, 10, 20}
$$\therefore P\left( E \right) = \frac{{n\left( E \right)}}{{n\left( S \right)}} = \frac{9}{{20}}$$
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LoginP(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
Why don't you guys just correct it.