A coin whose faces are marked 3 and 5 is tossed 4 times the probablity that the sum of the numbers thrown is greater than 15
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Answer:11/16
Step-by-step explanation:
Possible outcomes = {3, 5} i.e. 2 in number
As it is tossed four times, possible outcomes = 2*2*2*2 = 16
Favourable outcomes = {(5,5,5,5), (5,5,5,3), (5,5,3,5), (5,3,5,5), (3,5,5,5), (5,5,3,3), (5,3,3,5), (3,3,5,5), (3,5,5,3), (3,5,3,5), (5,3,5,3)} i.e. 11 in no.
Probability= 11/16
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