91Ó°ÊÓ

Here,\({p_i}\)be the probability of head occurring on the i-th coin toss, where\(i = (1,2,3,4,5)\) .

The probabilities of observing a head on different coins:\({p_1} = 0\),\({p_2} = \frac{1}{4}\),\({p_3} = \frac{1}{2}\),\({p_4} = \frac{3}{4}\)and\({p_5} = 1\).

Let us consider the events,

\({C_i}\)is the event that i-thcoin is tossed and\(H\)is that a head is obtained.

The probability that a coin is selected is, \(\Pr \left( {{C_i}} \right) = \frac{1}{5}\) and \({p_i} = \Pr \left( {{H_1}\left| {{C_i}} \right.} \right)\) .

a.

The probability of selectingi-th coin given that a head appears, is\(\Pr \left( {{C_i}\left| H \right.} \right)\).

\(\begin{aligned}{}{\bf{Pr}}\left( {{{\bf{C}}_{\bf{i}}}\left| {\bf{H}} \right.} \right){\bf{ = }}\frac{{{\bf{Pr}}\left( {{{\bf{C}}_{\bf{i}}}} \right){\bf{Pr}}\left( {{\bf{H}}\left| {{{\bf{C}}_{\bf{i}}}} \right.} \right)}}{{\sum\limits_{{\bf{j = 1}}}^{\bf{5}} {{\bf{Pr}}\left( {{{\bf{C}}_{\bf{j}}}} \right){\bf{Pr}}\left( {{\bf{H}}\left| {{{\bf{C}}_{\bf{j}}}} \right.} \right)} }}\\{\bf{ = }}\frac{{\frac{{\bf{1}}}{{\bf{5}}}{\bf{ \times }}{{\bf{p}}_{\bf{i}}}}}{{\frac{{{{\bf{p}}_{\bf{1}}}{\bf{ + }}{{\bf{p}}_{\bf{2}}}{\bf{ + }}{{\bf{p}}_{\bf{3}}}{\bf{ + }}{{\bf{p}}_{\bf{4}}}{\bf{ + }}{{\bf{p}}_{\bf{5}}}}}{{\bf{5}}}}}\\{\bf{ = }}\frac{{{{\bf{p}}_{\bf{i}}}}}{{{\bf{0 + }}\frac{{\bf{1}}}{{\bf{4}}}{\bf{ + }}\frac{{\bf{1}}}{{\bf{2}}}{\bf{ + }}\frac{{\bf{3}}}{{\bf{4}}}{\bf{ + 1}}}}\\{\bf{ = }}\frac{{{\bf{2}}{{\bf{p}}_{\bf{i}}}}}{{\bf{5}}}\end{aligned}\)

Thus, the probability that 1^st coin is selected given that the head appears on the toss is,

\(\begin{aligned}{}\Pr \left( {{C_1}\left| H \right.} \right) = \frac{{2 \times 0}}{5}\\ = 0\end{aligned}\)

Thus, the probability that 2^nd coin is selected given that the head appears on the toss is,

\(\begin{aligned}{}\Pr \left( {{C_2}\left| H \right.} \right) = \frac{{2\left( {\frac{1}{4}} \right)}}{5}\\ = \frac{1}{{10}}\end{aligned}\)

Thus, the probability that 3rd coin is selected given that the head appears on the toss is,

\(\begin{aligned}{}\Pr \left( {{C_3}\left| H \right.} \right) = \frac{{2\left( {\frac{1}{2}} \right)}}{5}\\ = \frac{1}{5}\end{aligned}\)

Thus, the probability that 3rd coin is selected given that the head appears on the toss is,

\(\begin{aligned}{}\Pr \left( {{C_4}\left| H \right.} \right) = \frac{{2\left( {\frac{3}{4}} \right)}}{5}\\ = \frac{3}{{10}}\end{aligned}\)

Thus, the probability that 3rd coin is selected given that the head appears on the toss is,

\(\begin{aligned}{}\Pr \left( {{C_5}\left| H \right.} \right) = \frac{{2\left( 1 \right)}}{5}\\ = \frac{2}{5}\end{aligned}\)

b.

It is known that the same coin where head appears is tossed again. Define,\({H_2}\)as the event that obtaining another head.

Thus, the required probability is\(\Pr \left( {{H_2}\left| H \right.} \right)\)computed using the following formulae,

\({\bf{Pr}}\left( {{{\bf{H}}_{\bf{2}}}\left| {\bf{H}} \right.} \right){\bf{ = }}\sum\limits_{{\bf{i = 1}}}^{\bf{5}} {{{\bf{p}}_{\bf{i}}}{\bf{ \times Pr}}\left( {{{\bf{C}}_{\bf{i}}}\left| {\bf{H}} \right.} \right)} \)

Therefore, the probability is obtained as,

\(\begin{aligned}{}\Pr \left( {{H_2}\left| H \right.} \right) = \left( {0 \times 0} \right) + \left( {\frac{1}{4} \times \frac{1}{{10}}} \right) + \left( {\frac{1}{2} \times \frac{1}{5}} \right) + \left( {\frac{3}{4} \times \frac{3}{{10}}} \right) + \left( {1 \times \frac{2}{5}} \right)\\ = 1 + \frac{1}{{40}} + \frac{1}{{10}} + \frac{9}{{40}} + \frac{2}{5}\\ = \frac{3}{4}\end{aligned}\)

Therefore, there is a\(\frac{3}{4}\)chance that the selected coin gives another head after giving a head in the first toss.

c.

It is given that the tail is observed on the selected coin in the first toss.

Let\({T_n}\)be the event that a tail is observed on the n^thtoss.

The probability of getting a tail from different coins is complementary to event of getting a head on the coin toss.

Therefore,

\(\begin{aligned}{}p_1^c = 1 - {p_1}\\ = 1 - 0\\ = 1\end{aligned}\)

Similarly, \(p_2^c = \frac{3}{4}\) ,\(p_3^c = \frac{1}{2}\) ,\(p_4^c = \frac{1}{4}\) and \(p_5^c = 0\)

The probability of selecting a coin if the randomly selected coin lands on a tail in the first toss, that is\(\Pr \left( {{C_i}\left| {{T_1}} \right.} \right)\).

Therefore,

\(\begin{aligned}{}{\bf{Pr}}\left( {{{\bf{C}}_{\bf{i}}}\left| {{{\bf{T}}_{\bf{1}}}} \right.} \right){\bf{ = }}\frac{{{\bf{Pr}}\left( {{{\bf{C}}_{\bf{i}}}} \right){\bf{Pr}}\left( {{{\bf{T}}_{\bf{1}}}\left| {{{\bf{C}}_{\bf{i}}}} \right.} \right)}}{{\sum\limits_{{\bf{j = 1}}}^{\bf{5}} {{\bf{Pr}}\left( {{{\bf{C}}_{\bf{j}}}} \right){\bf{Pr}}\left( {{{\bf{T}}_{\bf{1}}}\left| {{{\bf{C}}_{\bf{j}}}} \right.} \right)} }}\\ = \frac{{\frac{1}{5} \times p_i^c}}{{\frac{{p_1^c + p_2^c + p_3^c + p_4^c + p_5^c}}{5}}}\\ = \frac{{2p_i^c}}{5}\end{aligned}\)

Thus, given there is a tail in first toss, the probability of selecting the first coin is,\(\Pr \left( {{C_1}\left| {{T_1}} \right.} \right) = \frac{2}{5}\)

Similarly, for,\(\Pr \left( {{C_2}\left| {{T_1}} \right.} \right) = \frac{3}{{10}}\), \(\Pr \left( {{C_3}\left| {{T_1}} \right.} \right) = \frac{1}{5}\), \(\Pr \left( {{C_4}\left| {{T_1}} \right.} \right) = \frac{1}{{10}}\) , \(\Pr \left( {{C_5}\left| {{T_1}} \right.} \right) = 0\)for second, third, fourth and fifth coin respectively.

Now the probability of obtaining a head in the second toss when a tail had been obtained in the first toss is,\(\Pr \left( {{H_2}\left| {{T_1}} \right.} \right)\).

Therefore,

\(\begin{aligned}{}{\bf{Pr}}\left( {{{\bf{H}}_{\bf{2}}}\left| {{{\bf{T}}_{\bf{1}}}} \right.} \right){\bf{ = }}\sum\limits_{{\bf{i = 1}}}^{\bf{5}} {{\bf{Pr}}\left( {{{\bf{H}}_{\bf{2}}}\left| {{{\bf{C}}_{\bf{i}}}} \right.} \right){\bf{ \times Pr}}\left( {{{\bf{C}}_{\bf{i}}}\left| {{{\bf{T}}_{\bf{1}}}} \right.} \right)} \\ = \sum\limits_{{\bf{i = 1}}}^{\bf{5}} {{{\bf{p}}_{\bf{i}}}{\bf{ \times Pr}}\left( {{{\bf{C}}_{\bf{i}}}\left| {{{\bf{T}}_{\bf{1}}}} \right.} \right)} \\ = \left( {0 \times \frac{2}{5}} \right) + \left( {\frac{1}{4} \times \frac{3}{{10}}} \right) + \left( {\frac{1}{2} \times \frac{1}{5}} \right) + \left( {\frac{1}{{10}} \times \frac{3}{4}} \right) + \left( {1 \times 0} \right)\\ = \frac{1}{4}\end{aligned}\)

Thus, there are 25% chance of getting a head in the second toss given that we had obtained a tail in the first toss.