Sum
A box contains 36 cards, bearing only one number from 1 to 36 on each. If one card is drawn at random, find the probability of an event that the card drawn bears, a number divisible by 3
Advertisement Remove all ads
Solution
Sample space,
S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10,
11, 12, 13, 14, 15, 16, 17, 18, 19, 20,
21, 22, 23, 24, 25, 26, 27, 28, 29, 30,
31, 32, 33, 34, 35, 36}
∴ n(S) = 36
Let C be the event that the card drawn bears a number divisible by 3.
∴ C = {3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36}
∴ n(C) = 12
∴ P(C) = `("n"("C"))/("n"("S"))`
∴ P(C) = `12/36`
∴ P(C) = `1/3`
Concept: Probability of an Event
Is there an error in this question or solution?
APPEARS IN
Advertisement Remove all ads