Question

Suppose that a box contains one fair coin and one coin with a head on each side. Suppose that a coin is drawn at random from this box and that we begin to flip the coin. In Eqs. (2.3.4) and (2.3.5), we computed the conditional probability that the coin was fair, given that the first two flips both produce heads.

a. Suppose that the coin is flipped a third time and another head is obtained. Compute the probability that the coin is fair, given that all three flips produced

heads.

b. Suppose that the coin is flipped a fourth time, and the result is tails. Compute the posterior probability that the coin is fair.

Short answer

a. The coin's probability is fair given that all three flips obtained head is \(\frac{1}{9}\).

b. The posterior probability that the coin is fair, given that there was a tail at the fourth flip, is 1

Step-by-step solution

Given information

A box contains two coins. One is fair, and another one is unfair; it has a head on both sides. Out of them, one is randomly drawn from the box and begins to flip.

Now referring to Eqs. (2.3.4) and Eqs. (2.3.5), we obtain the conditional probability that the randomly drawn coin was fair given that there are two heads on the first two flips. And the probability is \(\frac{1}{5}\).

State the events

Consider the events,

\({B_1} = \)The event that the drawn coin is fair

\({B_2} = \)The event that the drawn coin has two heads on both sides

\({H_n} = \)The event of head obtained on the n^th flip

Calculate the probability

Referring to Eqs. (2.3.4) and Eqs. (2.3.5), we got the probability

\(\begin{aligned}{}{\bf{Pr}}\left( {{{\bf{B}}_{\bf{1}}}\left| {{{\bf{H}}_{\bf{1}}} \cap {{\bf{H}}_{\bf{2}}}} \right.} \right){\bf{ = }}\frac{{{\bf{Pr}}\left( {{{\bf{B}}_{\bf{1}}}\left| {{{\bf{H}}_{\bf{1}}}} \right.} \right){\bf{Pr}}\left( {{{\bf{H}}_{\bf{2}}}\left| {{{\bf{B}}_{\bf{1}}} \cap {{\bf{H}}_{\bf{1}}}} \right.} \right)}}{{{\bf{Pr}}\left( {{{\bf{B}}_{\bf{1}}}\left| {{{\bf{H}}_{\bf{1}}}} \right.} \right){\bf{Pr}}\left( {{{\bf{H}}_{\bf{2}}}\left| {{{\bf{B}}_{\bf{1}}} \cap {{\bf{H}}_{\bf{1}}}} \right.} \right){\bf{ + Pr}}\left( {{{\bf{B}}_{\bf{2}}}\left| {{{\bf{H}}_{\bf{1}}}} \right.} \right){\bf{Pr}}\left( {{{\bf{H}}_{\bf{2}}}\left| {{{\bf{B}}_{\bf{2}}} \cap {{\bf{H}}_{\bf{1}}}} \right.} \right)}}\\ = \frac{1}{5}\end{aligned}\)

a.

Now the required probability that the selected coin is fair, given that the coin gives another head-on 3^rd flip after giving two heads is,

\(\begin{aligned}{}{\bf{Pr}}\left( {{{\bf{B}}_{\bf{1}}}\left| {{{\bf{H}}_{\bf{1}}} \cap {{\bf{H}}_{\bf{2}}} \cap {{\bf{H}}_{\bf{3}}}} \right.} \right){\bf{ = }}\frac{{{\bf{Pr}}\left( {{{\bf{B}}_{\bf{1}}}\left| {{{\bf{H}}_{\bf{1}}} \cap {{\bf{H}}_{\bf{2}}}} \right.} \right){\bf{Pr}}\left( {{{\bf{H}}_{\bf{3}}}\left| {{{\bf{B}}_{\bf{1}}} \cap {{\bf{H}}_{\bf{1}}} \cap {{\bf{H}}_{\bf{2}}}} \right.} \right)}}{{\left[ \begin{aligned}{}{\bf{Pr}}\left( {{{\bf{B}}_{\bf{1}}}\left| {{{\bf{H}}_{\bf{1}}} \cap {{\bf{H}}_{\bf{2}}}} \right.} \right){\bf{Pr}}\left( {{{\bf{H}}_{\bf{3}}}\left| {{{\bf{B}}_{\bf{1}}} \cap {{\bf{H}}_{\bf{1}}} \cap {{\bf{H}}_{\bf{2}}}} \right.} \right){\bf{ + }}\\{\bf{Pr}}\left( {{{\bf{B}}_{\bf{2}}}\left| {{{\bf{H}}_{\bf{1}}} \cap {{\bf{H}}_{\bf{2}}}} \right.} \right){\bf{Pr}}\left( {{{\bf{H}}_{\bf{3}}}\left| {{{\bf{B}}_{\bf{2}}} \cap {{\bf{H}}_{\bf{1}}} \cap {{\bf{H}}_{\bf{2}}}} \right.} \right)\end{aligned} \right]}}\\ = \frac{{\frac{1}{5} \times \frac{1}{2}}}{{\left( {\frac{1}{5} \times \frac{1}{2}} \right) + \left( {\frac{4}{5} \times \frac{1}{2}} \right)}}\\ = \frac{1}{9}\end{aligned}\)

Thus, the probability is \(\frac{1}{9}\)

Calculate the probability

b.

The randomly drawn same coin is flipped for the fourth time and given that the result is a tail.

We have to find the posterior probability that the coin is fair.

Now we know that there are only two coins in the box: one fair coin and another coin with two heads on both sides. As there is a tail in the result, the coin with two heads on both sides cannot be drawn at first.

Thus, we can conclude that the posterior probability of the drawn coin being fair is 1.