A = P(1 + rt/100)
So, A= 3P (it becomes triples)
3P = P(1 + 8r/100)?r = 25%
Now, when t = 20
?A = P(1 + 25*20/100)
?A = P(1+ 5)?A = 6P
So, it becomes 6 times.
Possible outcomes = (RY, YP, PR)
2C1 4C1 + 4C1 6C1 + 6C1 2C1
Required probability = (2C1 4C1 + 4C1 6C1 + 6C1 2C1)/(12C2)
= 8 + 24 + 12/66
= 44/66
= 2/3.
Here S = {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH}
Let E = event of getting at most two heads.
Then E = {TTT, TTH, THT, HTT, THH, HTH, HHT}.
P(E) =n(E)/n(S)=7/8.
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