A box contains 36 tickets numbered 1 to 36. 1 ticket drawn at random. Find the probability that the number on the ticket is either divisible by 3 or is a perfect square.
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1/3 is the probability of the ticket being divisible by 3 and of it being a prime number is 11/36
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P(number on ticket that is divisible by 3 or a perfect square)=2/36=1/18
since 9 and 36 are the only possibilities
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