Probability Question 114
Question: A coin is tossed 2n times. The chance that the number of times one gets head is not equal to the number of times one gets tail is
[DCE 2002]
Options:
A) $ \frac{(2n!)}{{{(n!)}^{2}}}{{( \frac{1}{2} )}^{2n}} $
B) $ 1-\frac{(2n!)}{{{(n!)}^{2}}} $
C) $ 1-\frac{(2n!)}{{{(n!)}^{2}}}.\frac{1}{4^{n}} $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
The required probability = 1 - Probability of equal number of heads and tails
$ =1-{{}^{2n}}C_{n}{{( \frac{1}{2} )}^{n}}{{( \frac{1}{2} )}^{2n-n}}=1-\frac{(2n)!}{n!n!}{{( \frac{1}{4} )}^{n}}=1-\frac{(2n)!}{{{(n!)}^{2}}}\cdot \frac{1}{4^{n}} $ .
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