Solved Example #2

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 = 2^42. Thus,Option C

Solved Example #2
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