Question

A biased coin with probability p$;$ 0$<$p$<$1 of head 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 equals to

• A. $\frac{1}{5}$
• B. $\frac{2}{3}$
• C. $\frac{2}{5}$
• D. $\frac{3}{5}$

Answer 1

Answer: (D)

$\frac{3}{5}$

Answer 2

Let's denote the probability of getting a head on a single toss as p, where 0$<$p$<$1.
The probability of getting a tail on a single toss is then q = 1- p
Now, to find the probability of needing an even number of tosses to get the first head, we can consider the following:
1. You can get a head on the first toss (probability = p).
2. You can get a tail on the first toss (probability = q) and then get a head on the second toss (probability = p).
The total probability of needing an even number of tosses is the sum of these two cases:
P(even number of tosses) = p$+$(q$\times$p)
Now, we're given that P(even number of tosses) =$\frac{2}{5}$. So:
p$+$(q$\times$p) =$\frac{2}{5}$
Now, substitute q = 1- p:
p$+$((1- p)$\times$p) =$\frac{2}{5}$
Now, solve for p:
p$+$($p$- $p^2$) =$\frac{2}{5}$
Combine like terms: 2p- $p^2$ =$\frac{2}{5}$
Multiply both sides by 5 to get rid of the fraction:
10p-5$p^2$=2
Rearrange the terms:
5$p^2$-10p$+$2=0
Now, we can solve this quadratic equation for p.
We can use the quadratic formula:
$p=-b\underline+\sqrt\frac{(b2-4ac)}{2a}$
In this case, a=5,b=-10,andc=2.
$p=-\frac{(-10)\underline+\sqrt((-10)2-4(5)(2))}{2(5)}$
$p=10\underline+\sqrt\frac{(100-40)}{10}$
$p=10\underline+\sqrt\frac{60}{10}$
Now, simplify:
$p=10\underline+2\sqrt\frac{15}{10}$
Factor out 2 from the numerator:
$p=\frac{2(5\underline+\sqrt15)}{10}$
Now, cancel out the common factor of 2 in the numerator and denominator:
$p=5\underline+\sqrt\frac{15}{5}$
So, the possible values of pare:

1. $p=5+\sqrt\frac{15}{5}$
2. $p=5-\sqrt\frac{15}{5}$
   Since0$<$p$<$1, the correct value of p is
   $p=5-\sqrt\frac{15}{5}$
   So, the correct option is (D): $\frac{3}{5}$.