Probability Question 369
Question: A fair coin is tossed a fixed number of times. If the probability of getting seven heads is equal to that of getting nine heads, the probability of getting two heads is
Options:
A) $ \frac{15}{2^{8}} $
B) $ \frac{2}{15} $
C) $ \frac{15}{2^{13}} $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
Let n be the number of tosses and X the number of times heads occurs.
Then $ X\tilde{\ }B(n,p), $ with $ p=1/2 $ .
Therefore, since $ P(X=7)=P(X=9), $
we have $ ^{n}C_7{{( \frac{1}{2} )}^{7}}{{( \frac{1}{2} )}^{n-1}}{{=}^{n}}C_9{{( \frac{1}{2} )}^{9}}{{( \frac{1}{2} )}^{n-9}}{{\Rightarrow }^{n}}C_7{{( \frac{1}{2} )}^{n}}{{=}^{n}}c_9{{( \frac{1}{2} )}^{n}} $
That is, $ ^{n}C_7{{=}^{n}}C_9{{=}^{n}}{C_{n-9}}, $ yielding $ 7=n-9 $ or $ n=16. $
Hence $ P(X=2){{=}^{16}}C_2{{( \frac{1}{2} )}^{16}}=( \frac{16\times 15}{2} ){{( \frac{1}{2} )}^{16}}=\frac{15}{2^{13}} $
BETA
