Probability Question 309
Question: A fair coin is tossed a fixed number of times. If the probability of getting 7 heads is equal to that of getting 9 heads, then the probability of getting 3 heads is
Options:
A) $ \frac{35}{2^{12}} $
B) $ \frac{35}{2^{14}} $
C) $ \frac{7}{2^{12}} $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
Let the coin be tossed $ n $ times $ P $ (7 heads) $ ={}^{n}C_7{{( \frac{1}{2} )}^{7}}{{( \frac{1}{2} )}^{n-7}}={}^{n}C_7{{( \frac{1}{2} )}^{n}} $ and $ P $ (9 heads) $ ={}^{n}C_9{{( \frac{1}{2} )}^{9}}{{( \frac{1}{2} )}^{n-9}}={}^{n}C_9{{( \frac{1}{2} )}^{n}} $
$ P $ (7 heads) $ =P $ (9 heads)
$ \Rightarrow {}^{n}C_7={}^{n}C_9\Rightarrow n=16 $
$ \therefore P $ (3 heads) $ ={}^{16}C_3{{( \frac{1}{2} )}^{3}}{{( \frac{1}{2} )}^{16-3}}={}^{16}C_3{{( \frac{1}{2} )}^{16}}=\frac{35}{2^{12}}. $
BETA
