Question

Independent trials, each resulting in success with probability \(p\), are performed. (a) Find the expected number of trials needed for there to have been both at least \(n\) successes or at least \(m\) failures. Hint: Is it useful to know the result of the first \(n+m\) trials? (b) Find the expected number of trials needed for there to have been either at least \(n\) successes or at least \(m\) failures. Hint: Make use of the result from part (a).

Short answer

The expected number of trials needed to achieve either at least \(n\) successes or at least \(m\) failures is given by: \[E(k) = \sum_{k=0}^\infty k \sum_{i=0}^{n-1} \sum_{j=0}^{m-1}\binom{k}{i}\binom{k}{j}p^{i+j}(1-p)^{2k-i-j} \] To find the expected number of trials, plug in the specific values for \(p\), \(n\), and \(m\).

Step-by-step solution

Find the Probability

We want to find the probability \(\displaystyle P( X\geq n\ \text{or}\ Y\geq m)\), where \(\displaystyle X\) represents the number of successes and \(\displaystyle Y\) represents the number of failures. Define \(\displaystyle A\) as the event that there are at least \(\displaystyle n\) successes in a fixed number of trials, and \(\displaystyle B\) as the event that there are at least \(\displaystyle m\) failures in the same number of trials. Using complementary probabilities, we can write the desired probability as: \(\displaystyle P( A\ \text{or}\ B)=1-P( A^{C} \ \text{and}\ B^{C})\) where \(\displaystyle A^{C}\) denotes the complementary event of \(\displaystyle A\) (not \(\displaystyle A\) happening) and similarly for \(\displaystyle B^{C}\). Since the trials are independent, we have: \(\displaystyle P( A^{C} \ \text{and}\ B^{C})= P( A^{C})P( B^{C})\) The problem ultimately asks us to compute the expected number of trials needed for either \(\displaystyle A\) or \(\displaystyle B\) to happen.

Calculate the Expected Number of Trials

The expected value can be computed as the sum of probabilities for reality. Let \(\displaystyle k\) be the number of trials; then the expected number of trials can be calculated as: \[E(k) = \sum_{k=0}^{\infty} k \cdot P_k(A \ \text{or} \ B) \] Substituting the complementary probability, \[E(k) = \sum_{k=0}^\infty k[P(A^{C} \ \text{and} \ B^{C})_k] \] Since the trials are independent, we can write \[ P(A^C) = \left( 1-\sum_{i=0}^{n-1} \binom{k}{i} p^i(1-p)^{k-i} \right) \] and \[ P(B^C) = \left( 1-\sum_{i=0}^{m-1} \binom{k}{i} p^{k-i}(1-p)^{i} \right) \] Substituting back, we get \[E(k) =\sum_{k=0}^\infty k \left [ \left( 1-\sum_{i=0}^{n-1} \binom{k}{i} p^i(1-p)^{k-i} \right)\left( 1-\sum_{i=0}^{m-1} \binom{k}{i} p^{k-i}(1-p)^{i} \right) \right ] \] Now, it's easier to manage tabs:

Simplify the expression

Using linearity of expectation, we have : \[E(k) = \sum_{k=0}^\infty k \sum_{i=0}^{n-1} \sum_{j=0}^{m-1}\binom{k}{i}\binom{k}{j}p^{i+j}(1-p)^{2k-i-j} \] The expression inside the summation represents the probability of obtaining exactly \(\displaystyle i\) successes and \(\displaystyle j\) failures. This is the expected number of trials needed to achieve either at least \(\displaystyle n\) successes or at least \(\displaystyle m\) failures. To calculate the value, plug in the specific values for \(\displaystyle p\), \(\displaystyle n\), and \(\displaystyle m\).