A box contains 49 tickets numbered 1 to 49. One ticket drawn at randomly, find the probability that number on the ticket is either divisible by 3 or is a perfect square?
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7
All possibilities out of a total of 49possibilities are —1, 4, 9, 16, 25, 36, 49, 8, 12, 20, 24, 28, 32, 40, 44, 48
That means a total of 16 positive outcomes.
Therefore probability will be 16/49
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Possible outcomes that it is a multiple of 3 = 3 , 6 , 9 , 12 , 15 , 18 , 21 , 24 , 27 , 30 , 33 , 36 , 39 , 42 , 45 , 48 . i. e , 16 .
Possible outcomes that it is perfect square = 1 , 4 , 9 , 16 , 25 , 36 , 49 . i.e , 7 .
Total possible outcomes = 16 + 7 = 23 .
The bolded numbers have occurred 2 times , so subtracting 2 from possible outcomes we get ,
= 23 - 2 = 21
Total outcomes = 49
Probability = Possible outcomes / Total outcomes
= 21 / 49
= 3 / 7
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