Question

A coin is randomly selected from a group of ten coins, the \(n\) th coin having a probability \(n / 10\) of coming up heads. The coin is then repeatedly flipped until a head appears. Let \(N\) denote the number of flips necessary. What is the probability distribution of \(N\) ? Is \(N\) a geometric random variable? When would \(N\) be a geometric random variable; that is, what would have to be done differently? You are invited to a party. Suppose the times at which invitees are independent

Short answer

The probability distribution of \(N\), the number of flips necessary for a head to appear, is given by: \[P(N=n) = \sum_{k=1}^{10}\left( (1-\frac{k}{10})^{n-1}\frac{k}{10}\times \frac{k}{10}\right)\] \(N\) is not a geometric random variable in this case, as it doesn't follow the geometric distribution formula. \(N\) would be a geometric random variable if all coins had the same probability of coming up heads, such as each having a probability of \(\frac{1}{2}\).

Step-by-step solution

Calculate the probability of selecting each coin

First, we find the probability of selecting each coin \(k\) in the group of ten coins. We know that the probability of selecting the \(k^{th}\) coin is \(\frac{k}{10}\), since the selection process is random. To find the probability of needing exactly \(n\) flips to get heads with the \(k^{th}\) coin, we can calculate the probability using the binomial formula.

Calculate the probability of N flips for each coin

Given that we have already selected the coin \(k\), the probability of obtaining heads on the \(n^{th}\) flip can be calculated as \(P(N=n|X=k)=(1-\frac{k}{10})^{n-1}\frac{k}{10}\), where \(n\) denotes the number of flips and \(X=k\) represents the event of selecting the coin \(k\). This formula takes into account that the first \(n-1\) flips should be tails and the final flip should be heads.

Calculate the overall probability distribution of N

To find the overall probability distribution of \(N\), we need to consider all possible coin selections. So, we calculate the probability for each value of \(n\) as follows: \[P(N=n) = \sum_{k=1}^{10}P(N=n|X=k)P(X=k) = \sum_{k=1}^{10}\left((1-\frac{k}{10})^{n-1}\frac{k}{10}\times \frac{k}{10}\right)\]

Determine if N is a geometric random variable

To check if \(N\) is a geometric random variable, we need to verify if the probability distribution follows the formula: \(P(N=n)=(1-p)^{n-1}p\), where \(p\) is a constant probability of success (getting heads in this case). However, since the probability distribution of \(N\) depends on the sum of probabilities from all the different coins, it doesn't follow the geometric distribution formula. Therefore, \(N\) is not a geometric random variable in this case.

Determine when N would be a geometric random variable

For \(N\) to be a geometric random variable, all coins must have the same probability of coming up heads. In other words, if the probability of getting heads is the same for all the coins, say \(p\), then the probability distribution of \(N\) would follow the geometric distribution formula: \(P(N=n)=(1-p)^{n-1}p\). This can be achieved by setting all the probabilities of getting heads to the same value, for instance, if each coin had a probability of \(\frac{1}{2}\) of coming up heads.