A hat contains 4 coins, where 3 are fair coins ( equally likely to land heads or

Question

A hat contains 4 coins, where 3 are fair coins ( equally likely to land heads or tails), but one is double-headed.

a) A cois is chosen uniformly at random and flipped. What is the probability that the coin lands heads?

b) Suppose I drew a coin at random and flipped it twice, and it landed heads-up 2 times in a row. What is the probability that I chose the double-headed coin?

c) How many times in a row would I have to flip heads in order to be 90% sure that I chose the double-headed coin?

d) How many times in a row would I have to flip heads in order to be 99% sure that I chose the double-headed coin?

Explanation / Answer

(a) Here a coin is chosen uniformy at random and flipped.

Pr(Coin lands heads) = Pr(the coin is among the three fair coins) * Pr(Head) + Pr(The coin is the two headed coin) * Pr(Head)

= 3/4 * 1/2 + 1/4 * 1 = 0.625

(b) Pr(2 head in two chances) = Pr(The coin is selected is fair one) * Pr(Two head in two toss) + Pr(The coin is the unfair one) * Pr(Two heads in two toss)

= 3/4 * 1/2 * 1/2 + 1/4 * 1 * 1 = 0.4375

Pr(The coin is two headed one l 2 head in two chanced) = 0.25/0.4375 = 4/7

(c) Here let say n is the number of time i have to flip heads to ensure 90% time that i chose a double headed coin.

Pr(n heads in n row from the selected coin) = 3/4 * (1/2)n + 1/4 * 1n = 1/4 (3/2n + 1)

Here

Pr(selected coin is two headed) = 1/4 / [1/4 (3/2n + 1]

now it is more than 90%

1/(3/2n + 1) = 0.90

3/2n + 1 = 1/0.90 = 1.1111

3/2n = 0.1111

2n = 27

n = 4.755

or n = 5

(d) Now we want to be 99% sure

so here

1/(3/2n + 1) = 0.99

3/2n + 1 = 1.0101

3/2n = 0.010101

2n = 297

n = 8.2143

or n = 9

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