Probability Question 211
Question: A biased coin with probability $ p,0<p<1, $ of heads is tossed until a head appears for the first time. If the probability that the number of tosses required is even is $ \frac{2}{5}, $ then $ p= $
Options:
A) $ \frac{1}{2} $
B) $ \frac{1}{3} $
C) $ \frac{1}{4} $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
Let $ X $ denotes the number of tosses required. Then $ P(X=r)={{(1-p)}^{r-1}}. $
$ p, $ for $ r=1,2,3…… $ Let $ E $ denote the event that the number of tosses required is even. Then $ P(E)=P[(X=2)\cup (X=4)\cup (X=6)\cup ……..] $
$ P(E)=P(X=2)+P(X=4)+P(X=6)+…… $
$ P(E)=(1-p)p+{{(1-p)}^{3}}p+{{(1-p)}^{5}}p+…….=\frac{1-p}{2-p} $ But we are given that $ P(E)=\frac{2}{5}, $ then we get $ p=\frac{1}{3}. $
BETA
