Advertisement Remove all ads

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

Advertisement Remove all ads
Advertisement Remove all ads
Advertisement Remove all ads
MCQ

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

Options

  • 1/3

  • 2/3

  • 2/5

  • 3/5

     
Advertisement Remove all ads

Solution

1/3
Probability of heads = p
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+ .....)

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

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

Concept: Bernoulli Trials and Binomial Distribution
  Is there an error in this question or solution?

APPEARS IN

RD Sharma Class 12 Maths
Chapter 33 Binomial Distribution
MCQ | Q 13 | Page 28

Video TutorialsVIEW ALL [1]

Advertisement Remove all ads
Share
Notifications

View all notifications


      Forgot password?
View in app×