How Is the Probability of the First Head on Odd Flips Calculated

[PMH_1]

Can anyone explain to me how this problem is solved

Determine the probability that the first head appears on an odd number of flips i.e. X contains {1,3,5..}.

P[X] = summation starting at x = 1 to infinity (1/2)^(2x-1) = 1/2 / (1 - 1/4) = 2/3

Basically my question is, how is the formula for P[X] obtained:
1. 1/2 / (1 - 1/4) : where does this come from?
2. summation starting at x = 1 to infinity (1/2)^(2x-1) and this as well

 

[mathman]

1. comes from 2. (sum of geometric series).

2. is derived from the fact that you need a sequence of an even number of tails followed by a head. Let n be number of tails P(n=0)=1/2, P(n=2)=1/8, P(n=4)=1/32. In general, P(n=2k)=(1/2)^(2k+1). Sum for k=0, infinity (since events are mutually exclusive and exhaustive). (your x=k+1).

 

[tacman]

.

The formula for P[X] is obtained using the geometric series formula, which states that the sum of an infinite geometric series with a common ratio r is equal to a / (1 - r), where a is the first term in the series and r is the common ratio. In this case, the first term is (1/2)^(2x-1) and the common ratio is (1/2)^2 = 1/4.

To understand where the formula comes from, let's break down the problem step by step. First, we need to understand what the event X represents. X contains all the possible outcomes where the first head appears on an odd number of flips. In other words, X contains all the possible outcomes where the first head appears on the first, third, fifth, and so on flips.

Now, let's consider each individual flip. The probability of getting a head on any given flip is 1/2, since we have a fair coin. This means that the probability of getting a head on the first flip is 1/2, the probability of getting a head on the third flip is also 1/2, and so on. Therefore, the probability of getting a head on the first, third, fifth, and so on flips is (1/2)^2, (1/2)^4, (1/2)^6, and so on. We can see that this forms a geometric series with a first term of (1/2)^2 and a common ratio of (1/2)^2.

Now, we can use the geometric series formula to calculate the sum of this infinite series. The first term is (1/2)^2 and the common ratio is (1/2)^2, so we have a = (1/2)^2 and r = (1/2)^2. Plugging these values into the formula, we get:

P[X] = (1/2)^2 / (1 - (1/2)^2) = (1/4) / (1 - 1/4) = (1/4) / (3/4) = 1/3

However, this is not the probability we are looking for. We want the probability of getting a head on an odd number of flips, which includes the first, third, fifth, and so on flips. So, we need to sum up the probabilities of all these