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Ticket numbered from 1 to 20 are mixed up and a ticket is drawn at random.
What is the probability that the ticket drawn has a number which is a multiple of 3 or 7 ?
multiples of 3
3 6 9 12 15 18
multiples of 7
all possible outcomes 6+2=8
From 1 to 20 numbers divisible by 3 are 6 by 7 are 2.
So, favourable outcomes are 8 out of a possible 20 outcomes
Probability = favourable outcomes / possible outcomes
= 8/20 = 2/5 = 0.4
p(E) = 8c1 / 20c1 = 8/20.....
multiples of 3= 3 , 6 , 9 , 12 ,15 , 18..
multiples of 7 = 7 , 14,,,,,
Tickets numbered 1 to 20 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn bears a number which is a multiple of 3.
A bag contains 5 white, 4 red and 3 black balls. A ball is drawn at random from the bag, what is the probability that it is not black?
A bag contain 5 blue and 4 black balls. Three balls are drawn at random. What is the probability that 2 are blue and 1 is black.
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