594 views
4 votes
4 votes
Consider a coin with probability P to be heads. What is the probability that the first head will appear on the even numbered tosses?

(A) (1 - P) / 2
(B) (1 - P) / (2 - P)
(C) (1 - P) / (3 - P)
(D) 1 / (P (1 - P))

2 Answers

Best answer
5 votes
5 votes

probability of head ==> p(h) = p               p(t) = (1-p)

 the first head will appear on the even numbered tosses so,

th, ttth , ttttth ............. infinite times

probability = (1-p)p + (1-p)^3 p +((1-p)^5)p+ ...........

acording to gp 

((1-p)p)/(1-((1-p)2

by solving above we get (1-p)/(2-p)

option b

selected by
1 votes
1 votes

P(heads) = P

P(tails) = 1-P

Coin is tossed even no of times means 2,4,6,8 ....... 

For any throw first head should appear after that tails on rest of the throws in individual trial.

P(E) = P(H)P(T) + P(H)P(T)P(T)P(T) + P(H)P(T)P(T)P(T)P(T)P(T)......

P(E)=$P(1-P) + P(1-P)^{3} + P(1-P)^{5} + ........ ( infinite\ GP)$

P(E) = $\frac{P(1-P)}{1-(1-P)^{2}} = \frac{P(1-P)}{(1-1+P)(1+1-P)} = \frac{1-P}{2-P}$

Hence option b) is correct

edited by

Related questions

0 votes
0 votes
0 answers
2
Debargha Mitra Roy asked Sep 26, 2023
175 views
Determine the geometric distribution for which the mean is 3 and variance is 4.
2 votes
2 votes
2 answers
4