Question

A coin is tossed an odd-number of times. The number of tosses is n where n = 2m + 1, and m
is a positive integer. Given, the probability of appearance of a head is p, and Pr[number of heads
exceeds number of tails by one) = Pr number of tails exceeds number of heads by three). Let X
denote the number of heads that appear.
Find p in terms of m. please solve it asap

Answer 1

Step-by-step explanation:

Let pnpn be the probability of getting an odd number of heads tossing a coin nn times, then

pn+1=(1−p1)⋅pn+p1(1−pn)pn+1=(1−p1)⋅pn+p1(1−pn)

where p1p1 is the probability to obtain a head with

Answer 2

Answer:

where p1 is the probability to obtain a head with one toss. So given p1 (if the coin is fair then p1=1/2), the above linear recursion allows us to evaluate pn for any positive integer n.

P.S. The recursion is explained as follows: at the (n+1)th toss we have an odd number of heads if and only if one of these disjoint cases occurs:

i) we have a tail and in the previous n tosses there are an odd number of heads;

ii) we have a head and in the previous n tosses there are an even number of heads.

Step-by-step explanation:

Hope this helps you and pls mark me as brainliest