Share

# 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 2/5, Then P Equals (A) 1/3 - Mathematics

ConceptBernoulli Trials and Binomial Distribution

#### 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 2/5, then p equals

• 1/3

• 2/3

• 2/5

• 3/5

#### Solution

1/3
Probability of Tails = (1 − p)
Probability that first head appears at even turn
Pe = (1 − p) p + (1 − p)3p+(1 − p)5p + .....
= (1 − p) p (1 + (1 − p)2 + (1 − p)4+ .....)

$= \left( 1 - p \right)p\left( \frac{1}{1 - \left( 1 - p \right)^2} \right)$
$= \left( 1 - p \right)p\left( \frac{1}{- p^2 + 2p} \right)$

$P_e = \frac{\left( 1 - p \right)}{\left( 2 - p \right)}$
$P_e = \frac{2}{5}$
$\frac{1 - p}{2 - p} = \frac{2}{5}$
$5 - 5p = 4 - 2p$
$3p = 1$
$p = \frac{1}{3}$

Is there an error in this question or solution?

#### Video TutorialsVIEW ALL [1]

Solution 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 2/5, Then P Equals (A) 1/3 Concept: Bernoulli Trials and Binomial Distribution.
S