Saturday, February 9th, 2013
Question: A fair coin is tossed 43 times. What is the number of cases where number of ‘Head’ > number of ‘Tail’?
(A) 2^43
(B) (2^43)-43
(C) 2^42
(D) None of the above.
Answer:
No. of heads > No. of tails
=> No. of heads can be 22, 23, 24… 43
Therefore  43C22 + 43C23 + 43C24 .. 43C43
We know that 43C0 + 43C1 + 43C2 … 43C43 = 2^43
We also know know that our answer has 22 terms which are 43C22 + 43C23 + 43C24 .. 43C43
These 22 terms are equal to = 43C0 + 43C1 + 43C2 .. 43C21
Both of them are equal and add up to 2^43
So each one of them is 1/2 of 2^43 = 2^42
43C22 + 43C23 + 43C24 .. 43C43 =Â 2^42. Thus,Option C
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