1. ## non-fair coin

Suppose that a non-fair coin with $P(\text{Heads}) = 0.6$ is tossed repeatedly.

(a) What is the probability that the first time we get "Heads" happens on an even numbered toss?

(b) What is the probability that the first "Heads" occurs on an odd-numbered toss?

(c) Explain why the answer (b) is larger than the answer in (a). Does it depend on $P(\text{Heads})$?

So let $P_{n}$ denote the probability that the first head appears on the $n$th toss. Then $P_{1} = 0.6, \ P_{2} = (0.6)(0.4), \ P(3) = (0.4)^{2}(0.6) \dots$

In general, $P_{n} = \frac{3}{5}\left(\frac{2}{5}\right)^{n-1}$. So for part (a) it would be $P_{2} + P_{4}+P_{6}+ \ldots + P_{2n}$ where $n \to \infty$.

Similarly, for part (b) it would be $P_{1}+P_{3} + P_{5} + \ldots + P_{2n+1}$ where $n \to \infty$.

So $P_{2n} = \frac{3}{5} \left(\frac{2}{5} \right)^{2n-1}$ and $P_{2n+1} = \frac{3}{5} \left(\frac{2}{5} \right)^{2n} = \frac{3}{5} \left(\frac{4}{25} \right)^{n}$.

Then $P_{2n+1} = \sum_{n=1}^{\infty} \frac{12}{125} \left(\frac{4}{25} \right)^{n-1} = \frac{\frac{12}{125}}{\frac{21}{25}} \approx 0.1142$. Then $P_{2n} = 1-0.1142 = 0.8858$?

It seems like $P_{2n+1} < P_{2n}$.

2. Originally Posted by lord12
Suppose that a non-fair coin with $P(\text{Heads}) = 0.6$ is tossed repeatedly.

(a) What is the probability that the first time we get "Heads" happens on an even numbered toss?

(b) What is the probability that the first "Heads" occurs on an odd-numbered toss?

(c) Explain why the answer (b) is larger than the answer in (a). Does it depend on $P(\text{Heads})$?

So let $P_{n}$ denote the probability that the first head appears on the $n$th toss. Then $P_{1} = 0.6, \ P_{2} = (0.6)(0.4), \ P(3) = (0.4)^{2}(0.6) \dots$

In general, $P_{n} = \frac{3}{5}\left(\frac{2}{5}\right)^{n-1}$. So for part (a) it would be $P_{2} + P_{4}+P_{6}+ \ldots + P_{2n}$ where $n \to \infty$.

Similarly, for part (b) it would be $P_{1}+P_{3} + P_{5} + \ldots + P_{2n+1}$ where $n \to \infty$.

So $P_{2n} = \frac{3}{5} \left(\frac{2}{5} \right)^{2n-1}$ and $P_{2n+1} = \frac{3}{5} \left(\frac{2}{5} \right)^{2n} = \frac{3}{5} \left(\frac{4}{25} \right)^{n}$.

Then $P_{2n+1} = \sum_{n=1}^{\infty} \frac{12}{125} \left(\frac{4}{25} \right)^{n-1} = \frac{\frac{12}{125}}{\frac{21}{25}} \approx 0.1142$. Then $P_{2n} = 1-0.1142 = 0.8858$?

It seems like $P_{2n+1} < P_{2n}$.
You seem to have lost $P_1$ somewhere.

$
P_{odd}=0.6 + 0.4^2 0.6 + ... =0.6 (1 + 0.4^2 + 0.4^4 ...)=0.6 \frac{1}{1-0.4^2} \approx 0.714..
$

RonL