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?
\displaystyle \frac{3^ k\cdot p}{3^{3-k}\cdot p+3^ k\cdot (1-p)}
\displaystyle \frac{p}{3^{3-k}}
\displaystyle \frac{3^{3-k}\cdot p}{3^{3-k}\cdot p+3^ k\cdot (1-p)}
\displaystyle \frac{3^{3-k}\cdot (1-p)}{3^{3-k}\cdot p+3^ k\cdot (1-p)}
unanswered
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.
k \leq \displaystyle \frac{3}{2}+\frac{1}{2}\log _3 \frac{p}{1-p}.
k \geq \displaystyle \frac{3}{2}+\frac{1}{2}\log _3 \frac{p}{1-p}.
k \leq \displaystyle \frac{1}{2}\log _3 \frac{p}{1-p}.
k \geq \displaystyle \frac{1}{2}\log _3 \frac{p}{1-p}.
none of the above
unanswered
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?
unanswered
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?
unanswered
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=\quad
\frac{1}{4}
To solve this problem, we will use Bayes' theorem to calculate the conditional probability that Bob received the first coin given that he observed k heads out of the 3 tosses.
Let A be the event that Bob received the first coin, and B be the event that Bob observed k heads out of the 3 tosses.
We are asked to find P(A|B), the probability that Bob received the first coin given that he observed k heads.
Bayes' theorem states that:
P(A|B) = (P(A) * P(B|A)) / P(B)
We are given that the probability of selecting the first coin is p, and the probability of selecting the second coin (1-p). We are also given the probabilities of getting a head for each coin.
P(A) = p (probability of selecting the first coin)
P(B|A) = (1/4)^k * (3/4)^(3-k) (probability of getting k heads given that Bob received the first coin)
Now, we need to calculate P(B), the probability of observing k heads.
P(B) = P(A) * P(B|A) + P(A') * P(B|A') (using the law of total probability)
P(B) = p * (1/4)^k * (3/4)^(3-k) + (1-p) * (3/4)^k * (1/4)^(3-k) (probability of getting k heads)
Now we can substitute these values into Bayes' theorem:
P(A|B) = (p * (1/4)^k * (3/4)^(3-k)) / (p * (1/4)^k * (3/4)^(3-k) + (1-p) * (3/4)^k * (1/4)^(3-k))
Simplifying this expression, we get:
P(A|B) = (3^k * p) / (3^(3-k) * p + 3^k * (1-p))
So the answer is \displaystyle \frac{3^ k\cdot p}{3^{3-k}\cdot p+3^ k\cdot (1-p)}.