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Suppose you are given a bag containing n unbiased coins. You are told that n − 1 of these coins are normal, with heads on one side and tails on the other, whereas one coin is a fake, with heads on both sides. . Suppose you reach into the bag, pick out a coin at random, flip it, and get a head. What is the (conditional) probability that the coin you chose is the fake coin?. Suppose you continue flipping the coin for a total of k times after picking it and see k heads. Now what is the conditional probability that you picked the fake coin? . Suppose you wanted to decide whether the chosen coin was fake by flipping it k times. The decision procedure returns fake if all k flips come up heads; otherwise it returns normal. What is the (unconditional) probability that this procedure makes an error?

Respuesta :

Answer:

a) [2/n+1]

b) 2ᵏ/[2ᵏ + (n-1)]

Explanation:

(n-1) coins are fair and only 1 is a fake coin with two heads.

This means there are a total of n coins in the bag.

Let the probability of getting a head be P(h)

Probability of picking a fake coin = P(f) = (1/n)

Probability of getting head on a fake coin = P(h|f) = 1

Probability of picking a fake coin and getting a head = P(f n h) = (1/n) × 1 = (1/n)

Probability of picking the right (original) coin = P(r) = (n-1)/n

Probability of getting a head on the right coin = P(h|r) = (1/2)

Probability of picking a real coin and getting a head = P(h n r) = [(n-1)/n] × (1/2) = [(n-1)/2n]

a) The required conditional probability is the probability that the coin is fake, given that it turns up head

P(f|h) = P(f n h)/P(h)

P(f n h) = (1/n)

But, we do not have P(h). We can obtain it through the relation,

P(h) = P(h n r) + P(h n f)

= P(r) P(h|r) + P(f) P(h|f)

= {[(n-1)/n] × [1/2] + [1/n] [1]

= [(n-1)/2n] + [1/n]

= (n+1)/2n

P(f|h) = P(f n h)/P(h)

= (1/n) ÷ [(n+1)/2n] = [2/n+1]

b) The second required probability

Probability that it's a fake coin, given that we get k heads.

P(f|k heads) = P(f n k heads) ÷ P(k heads)

P(f n k heads) = (1/n) × 1ᵏ = (1/n)

But, we do not have P(k heads). We can obtain it through the relation,

P(k heads) = P(r n k heads) + P(f n k heads)

P(r n k heads) = [(n-1)/n] × [1/2)ᵏ

= [(n-1)/2ᵏn]

P(f n k heads) = (1/n)

P(k heads) = P(r n k heads) + P(f n k heads)

= [(n-1)/2ᵏn] + [1/n]

P(f|k heads) = P(f n k heads) ÷ P(k heads)

= [1/n] ÷ {[(n-1)/2ᵏn] + [1/n]}

On simplifying,

We obtain

P(f|k heads) = 2ᵏ/[2ᵏ + (n-1)]

Note that, Bayes rule is the guiding principle in all the conditional probability evaluations we have done above.

Hope this Helps!!!

jayilych4real

The conditional probability that the coin you chose is the fake coin is:

  • a) [2/n+1]

The (unconditional) probability that this procedure makes an error is:

  • b) 2ᵏ/[2ᵏ + (n-1)]

To find the probability:

We know that(n-1) coins are fair and only 1 is a fake coin with two heads.

Therefore, there are a total of n coins in the bag.

Thus, the probability of getting a head be

  • P(h)

The probability of picking a fake coin =

  • P(f) = (1/n)

The probability of getting head on a fake coin =

  • P(h|f) = 1

The probability of picking a fake coin and getting a head =

  • P(f n h) = (1/n) × 1 = (1/n)

The probability of picking the right (original) coin =

  • P(r) = (n-1)/n

The probability of getting a head on the right coin =

  • P(h|r) = (1/2)

The probability of picking a real coin and getting a head =

  • P(h n r) = [(n-1)/n] × (1/2) = [(n-1)/2n]

Based on the conditional probability

  • P(f|h) = P(f n h)/P(h)
  • P(f n h) = (1/n)

= {[(n-1)/n] × [1/2] + [1/n] [1]

= [(n-1)/2n] + [1/n]

= (n+1)/2n

Hence, P(f|h) = P(f n h)/P(h)

= (1/n) ÷ [(n+1)/2n] = [2/n+1]

b) The second required probability

The probability that it's a fake coin, given that we get k heads is:

  • P(f|k heads) = P(f n k heads) ÷ P(k heads)
  • P(f n k heads) = (1/n) × 1ᵏ = (1/n)

  • P(k heads) = P(r n k heads) + P(f n k heads)
  • P(r n k heads) = [(n-1)/n] × [1/2)ᵏ

= [(n-1)/2ᵏn]

  • P(f n k heads) = (1/n)
  • P(k heads) = P(r n k heads) + P(f n k heads)

= [(n-1)/2ᵏn] + [1/n]

P(f|k heads) = P(f n k heads) ÷ P(k heads)

= [1/n] ÷ {[(n-1)/2ᵏn] + [1/n]}

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