Question

Suppose that a box contains five coins and that for each coin there is a different probability that a head will be obtained when the coin is tossed. Let \({{\bf{p}}_{\bf{i}}}\)denote the probability of a head when theith coin is tossed \({\bf{i = }}\left( {{\bf{1, \ldots ,5}}} \right)\) and suppose that \({{\bf{p}}_{\bf{1}}}{\bf{ = 0}}\),\({{\bf{p}}_{\bf{2}}}{\bf{ = }}\frac{{\bf{1}}}{{\bf{4}}}\) ,\({{\bf{p}}_{\bf{3}}}{\bf{ = }}\frac{{\bf{1}}}{{\bf{2}}}\) ,\({{\bf{p}}_{\bf{4}}}{\bf{ = }}\frac{{\bf{3}}}{{\bf{4}}}\) , and \({{\bf{p}}_{\bf{5}}}{\bf{ = 1}}\).

1. Suppose that one coin is selected at random from the box and when it is tossed once, a head is obtained. What is the posterior probability that theith coin was selected \({\bf{i = }}\left( {{\bf{1, \ldots ,5}}} \right)\)?
2. If the same coin were tossed again, what would be the probability of obtaining another head?
3. If a tail had been obtained on the first toss of the selected coin and the same coin were tossed again, what would be the probability of obtaining a head on the second toss?

Short answer

a. The posterior probability that the 1^st coin was selected is 0

The posterior probability that the 2^nd coin was selected is\(\frac{1}{{10}}\)

The posterior probability that the 3^rd coin was selected is\(\frac{1}{5}\)

The posterior probability that the 4^th coin was selected is\(\frac{3}{{10}}\)

The posterior probability that the 5^th coin was selected is\(\frac{2}{5}\)

b. If the same coin were tossed again, the probability of obtaining another head is \(\frac{3}{4}\)

c. If a tail had been obtained on the first toss of the selected coin and the same coin were tossed again, the probability of obtaining a head on the second toss is \(\frac{1}{4}\) .

Step-by-step solution

Given information

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\({p_5} = 1\).

State the events and probabilities

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 \({p_i} = \Pr \left( {{H_1}\left| {{C_i}} \right.} \right)\) .

Calculate the probability of selecting a coin

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,

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

Calculate the probability of getting consecutive heads

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.

Calculate the probability of getting head after tail

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.