91Ó°ÊÓ

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 \(\frac{1}{5}\).

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

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 \(\frac{1}{9}\)

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.