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

The Correct Option is D

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): \(\frac{3}{5}\).