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.
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)
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+1/2logp/1-p
For this part, assume that p=3/4.
What is the probability that Bob will guess the coin correctly using the decision rule from part 2?
Ans: 0.84375
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 : 0.75
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.
Ans : 0.03571
Why did the coin go to therapy? Because it had a lot of heads and tails.
To calculate the conditional probability that Bob received the first coin given k Heads out of the 3 tosses, we can use Bayes' theorem. The formula is:
P(First Coin | k Heads) = (P(k Heads | First Coin) * P(First Coin)) / P(k Heads)
Let's calculate each term:
P(k Heads | First Coin): Since the probability of getting a Head for the first coin is 1/4, the probability of getting k Heads out of 3 tosses is (1/4)^k * (3/4)^(3-k).
P(First Coin): The probability that Alice sends Bob the first coin is p, as given.
P(k Heads): To calculate this, we need to consider both the probabilities of getting k Heads with the first coin and with the second coin and their corresponding probabilities of getting selected by Alice.
P(k Heads) = P(k Heads | First Coin) * P(First Coin) + P(k Heads | Second Coin) * P(Second Coin)
= (1/4)^k * (3/4)^(3-k) * p + (3/4)^k * (1/4)^(3-k) * (1-p)
Now substituting all the values in the formula, we get:
P(First Coin | k Heads) = ((1/4)^k * (3/4)^(3-k) * p) / ((1/4)^k * (3/4)^(3-k) * p + (3/4)^k * (1/4)^(3-k) * (1-p))
= ((1/4)^k * (3/4)^(3-k) * p) / (3^k * (1-p) + (1-p))
Therefore, the conditional probability that Bob received the first coin given k Heads out of the 3 tosses is:
P(First Coin | k Heads) = (3^k * p) / (3^k * p + (1-p))
For the second part, we need to find the threshold condition that Bob should use to minimize the probability of error. The correct threshold condition is when k <= 3/2 + 1/2 * log(p / (1-p)).
To calculate the probability that Bob will guess the coin correctly using this decision rule, we need to calculate the probability that k satisfies the threshold condition. Assuming p=3/4, we can substitute this value in the threshold condition formula and check for each value of k:
k <= 3/2 + 1/2 * log(3/4 / (1-3/4))
By using this formula, we can calculate the probability that Bob will guess the coin correctly.
For the third part, where Bob guesses the coin without tossing it, we can simply use the probability that Alice selects the first coin, which is p. Therefore, the probability that Bob will guess the coin correctly is p.
For the last part, if p is small, then Bob will always decide in favor of the second coin, ignoring the results of the three tosses. So we need to find the maximum value of p where Bob will still decide in favor of the first coin.
Using the threshold condition formula from part 2:
t = maximum value of p where k <= 3/2 + 1/2 * log(p / (1-p))
By substituting the value of k = 3/2, we can solve for t.