Alice has two coins. The probability of Heads for the first coin is 1/4, and the probability of Heads for the second is 3/4. Other than this difference, the coins are indistinguishable. Alice chooses one of the coins at random and sends it to Bob. The random selection used by Alice to pick the coin to send to Bob is such that the first coin has a probability p of being selected. Assume that 0<p<1. Bob tries to guess which of the two coins he received by tossing it 3 times in a row and observing the outcome. Assume that for any particular coin, all tosses of that coin are independent.
1. Given that Bob observed k Heads out of the 3 tosses (where k=0,1,2,3), what is the conditional probability that he received the first coin?
ans= 3^3−k*p/3^3−k*p+3^k*(1−p)
2. We define an error to have occurred if Bob decides that he received one coin from Alice, but he actually received the other coin. He decides that he received the first coin when the number of Heads, k, that he observes on the 3 tosses satisfies a certain condition. When one of the following conditions is used, Bob will minimize the probability of error. Choose the correct threshold condition.
ans= k≤3/2+1/2log3*p/1−p.
3. For this part, assume that p=3/4.
(a) What is the probability that Bob will guess the coin correctly using the decision rule from part 2?
ans= unanswered
(b) Suppose instead that Bob tries to guess which coin he received without tossing it. He still guesses the coin in order to minimize the probability of error. What is the probability that Bob will guess the coin correctly under this scenario?
ans= unanswered
4. Bob uses the decision rule of Part 2. If p is small, then Bob will always decide in favor of the second coin, ignoring the results of the three tosses. The range of such p's is [0,t). Find t.
t=
ans =unanswered
To answer the remaining questions, let's break them down step by step:
3. For this part, assume that p=3/4.
(a) To find the probability that Bob will guess the coin correctly using the decision rule from part 2, we need to substitute p=3/4 into the condition k ≤ 3/2 + 1/2 * log₃(p/1−p).
- k ≤ 3/2 + 1/2 * log₃(3/4 / (1/4)) = 3/2 + 1/2 * log₃(3) ≈ 3/2 + 1/2 * 0.6309 ≈ 3/2 + 0.3154 ≈ 1.8154
Therefore, the condition becomes k ≤ 1.8154.
Since k can only take values from 0 to 3, the probability that Bob will guess the coin correctly is the probability of k being either 0, 1, 2, or 3, and satisfying the condition k ≤ 1.8154.
P(Bob guesses correctly) = P(k = 0) + P(k = 1) + P(k = 2) + P(k = 3), where k follows a binomial distribution with n = 3 and p = 3/4.
P(Bob guesses correctly) = C(3, 0) * (3/4)⁰ * (1 - 3/4)³ + C(3, 1) * (3/4)¹ * (1 - 3/4)² + C(3, 2) * (3/4)² * (1 - 3/4)¹ + C(3, 3) * (3/4)³ * (1 - 3/4)⁰
P(Bob guesses correctly) = (1 * 1 * 1/64) + (3 * 3/4 * 1/16) + (3 * 9/16 * 1/4) + (1 * 27/64 * 1)
P(Bob guesses correctly) ≈ 1/64 + 27/64 + 27/64 + 27/64 ≈ 82/64 ≈ 1.28125
(b) Now let's calculate the probability that Bob will guess the coin correctly without tossing it, using the decision rule to minimize the probability of error.
Since Bob still guesses which coin he received in order to minimize the probability of error, the probability that he guesses correctly is the same as the conditional probability that he received the first coin, given by the solution to question 1.
Therefore, the probability that Bob will guess the coin correctly without tossing it is the conditional probability P(First Coin | k Heads), where k = 0, 1, 2, or 3.
Using the formula from question 1, we can substitute p = 3/4 and k = 0, 1, 2, or 3 to find the respective conditional probabilities.
4. Bob uses the decision rule of Part 2. If p is small, then Bob will always decide in favor of the second coin, ignoring the results of the three tosses. The range of such p's is [0, t).
For Bob to always decide in favor of the second coin, the threshold condition k ≤ 3/2 + 1/2 * log₃(p/1−p) must always be false. In other words, the maximum value of k must be greater than 3/2 + 1/2 * log₃(p/1−p).
Since k can only take values from 0 to 3, the maximum value of k is 3. Therefore, we need to find the value of p that satisfies the inequality 3 > 3/2 + 1/2 * log₃(p/1−p).
Simplifying the inequality, we have 3 - 3/2 > 1/2 * log₃(p/1−p).
Multiplying both sides by 2, we get 6 - 9/2 > log₃(p/1−p).
Converting 6 into log₃, we have log₃(3²) - 9/2 > log₃(p/1−p).
Applying the logarithmic property logₐ(x * y) = logₐ(x) + logₐ(y), we get log₃(3²) - log₃(3⁹/₂) > log₃(p/1−p).
Using the logarithmic property logₐ(x^n) = n * logₐ(x), we have 2 * log₃(3) - log₃(3⁹/₂) > log₃(p/1−p).
Simplifying, we get 2 * 1 - log₃(3⁹/₂) > log₃(p/1−p).
Therefore, t is the solution to the inequality 2 - log₃(3⁹/₂) > log₃(p/1−p).
Calculating the value for t:
t = 2 - log₃(3⁹/₂) ≈ 1.7549
Answer for part 3(a) is 1.28125, and for part 3(b) and 4, you need to substitute the values and calculate the results yourself.