Problem 49 - A First Course in Probability

Question

There are two misshapen coins in a box; their probabilities for landing on heads when they are flipped are, respectively, 4 and \(.7 .\) One of the coins is to be randomly chosen and flipped 10 times. Given that two of the first three flips landed on heads, what is the conditional expected number of heads in the 10 flips?

Step-by-Step Solution

Answer

The conditional expected number of heads in the 10 flips is 5.85.

Step 1: Determine the probability of selecting each coin

There are two coins in the box, which means the probability of selecting each coin is the same: \(P(Coin1) = P(Coin2) = 0.5\)

Step 2: Calculate the probability of getting two heads in the first three flips for both coins

For Coin1: \(P(2H, 3F | Coin1) = \binom{3}{2}(0.4)^2(0.6)\) For Coin2: \(P(2H, 3F | Coin2) = \binom{3}{2}(0.7)^2(0.3)\)

Step 3: Calculate the conditional probabilities of getting a specific number of heads in the 10 flips for both coins

For Coin1: We have two options: two heads in the first two flips with probability \(0.4^2\) or in the first and third flips with probability \(2*0.4*0.6*0.4\). For Coin2: We have two options here as well: two heads in the first two flips with probability \(0.7^2\) or in the first and third flips with probability \(2*0.7*0.3*0.7\). We also have to calculate the conditional probabilities of getting a specific number of heads in the remaining seven flips. For Coin1, use a binomial probability distribution with \(n=7\), \(p=0.4\), and \(k\) ranging from 0 to 7. For Coin2, use a binomial probability distribution with \(n=7\), \(p=0.7\), and \(k\) ranging from 0 to 7.

Step 4: Calculate the conditional expected values of heads for both coins

For Coin1, the conditional expected value is given by: \(E[H | Coin1] = E[2 heads] + E[remaining 7 flips] = 2 + 0.4*7\) For Coin2, the conditional expected value is given by: \(E[H | Coin2] = E[2 heads] + E[remaining 7 flips] = 2 + 0.7*7\)

Step 5: Calculate the weighted average of the conditional expected values

The conditional expected number of heads for the entire experiment is found by taking the weighted sum of the expected values for each coin: \(E[H] = 0.5 * E[H|Coin1] + 0.5 * E[H|Coin2]\) After calculating the expected values and the weighted sum, we have: \(E[H] = 0.5*(2 + 0.4*7) + 0.5*(2 + 0.7*7) = 0.5*(4.8) + 0.5*(6.9) = 2.4 + 3.45 = 5.85\) So, the conditional expected number of heads in the 10 flips is 5.85.

Key Concepts

Conditional Expectation

Understanding conditional expectation is essential in problems involving probability, where some events have already occurred. Conditional expectation gives an expected value of a random variable given that a certain condition is met. It's denoted as \(E(X|Y)\), where \(X\) is the random variable and \(Y\) is the condition or event.
In the context of the coin-flipping problem, the condition is that two out of the first three flips land on heads. We want to find the expected number of heads in the total ten flips, conditioned on this initial outcome. This means calculating the expected number of heads for each coin after considering the condition.

• If you choose Coin1, which has a probability of 0.4 for heads, the conditional expectation is calculated over the remaining seven flips, using the probability of 0.4 per flip.
• Similarly, for Coin2, with a 0.7 probability, you use the same approach focusing on the remaining flips.

Conditional expectation allows us to incorporate known information to make more accurate predictions about unknown outcomes. This concept often teams up with other probability tools like Bayes' Theorem for comprehensive analyses.

Bayes' Theorem

Bayes' Theorem is a powerful tool in probability that allows us to update the probability of a hypothesis when given new evidence. It is particularly useful in decision-making and statistics, especially when we need to calculate conditional probabilities. The theorem is expressed as:\[ P(A|B) = \frac{P(B|A) \cdot P(A)}{P(B)} \]where:

• \(P(A|B)\) is the probability of event \(A\) occurring given event \(B\) has occurred.
• \(P(B|A)\) is the probability of event \(B\) given event \(A\) is true.
• \(P(A)\) and \(P(B)\) are the independent probabilities of events \(A\) and \(B\) respectively.

In the coin scenario, Bayes' Theorem is instrumental in determining which coin might have been chosen given that we observe two heads in the first three flips. By applying Bayes’ calculations, we derive how likely we are to have picked either coin under the condition of our preliminary observation. Once those probabilities are known, we can find out the expected number of heads across the remaining flips according to the correct probability distribution.

Binomial Distribution

The binomial distribution gives us the probability of achieving \(k\) successes in \(n\) trials, considering that each trial is independent and has a binary outcome (like flipping heads or tails). This distribution is likened to a series of Bernoulli trials. It is described by the formula:\[ P(X=k) = \binom{n}{k} p^k (1-p)^{n-k} \]where:

• \(\binom{n}{k}\) is the binomial coefficient, calculated as \(\frac{n!}{k!(n-k)!}\).
• \(p\) is the probability of a success, such as getting a head in each flip.
• \((1-p)\) is the probability of a failure.

In this exercise, the binomial distribution helps us calculate the probability of getting two heads in the first three flips and extends to calculating outcomes for the remaining trials. For Coin1, with \(p=0.4\), and Coin2, with \(p=0.7\), we evaluate the likelihood of various numbers of heads appearing. It’s integral to solving the problem accurately, as it defines the probability landscape from which expectations are calculated. By leveraging the entire distribution, we perform precise calculations of probabilities for multiple head counts.