Probability Question 398
Question: A fair coin it tossed 2n times. The probability of getting as many heads in the first n tosses as in the last n is
Options:
A) $ \frac{^{2n}C_{n}}{2^{2n}} $
B) $ \frac{^{2n}{C_{n-1}}}{2^{n}} $
C) $ \frac{n}{2^{2n}} $
D) None
Show Answer
Answer:
Correct Answer: A
Solution:
The number of possible outcomes of 2n tosses is $ 2^{2n}. $ There are $ ^{n}C_{r} $ ways of getting r heads, with $ 0\le r\le n, $ in n tosses.
Therefore, the number of ways of getting r heads in both the first n and last n tosses is $ {{{{(}^{n}}C_{r})}^{2}}. $
Summing over all values of r.
the number of favourable ways is $ {{{{(}^{n}}C_0)}^{2}}+{{{{(}^{n}}C_1)}^{2}}+{{{{(}^{n}}C_2)}^{2}}+…+{{{{(}^{n}}C_{n})}^{2}}{{=}^{2n}}C_{n}, $
So that the required probability is $ \frac{^{2n}C_{n}}{2^{2n}}. $
BETA
