Question

Consider again the box containing the five different coins described in Exercise 7. Suppose that one coin is selected at random from the box and is tossed repeatedly until a head is obtained.

a. If the first head is obtained on the fourth toss, what is the posterior probability that theith coin was selected (i=1, . . .,5)?

b. If we continue to toss the same coin until another head is obtained, what is the probability that exactly three additional tosses will be required?

Short answer

a. If the first head is obtained on the fourth toss,

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

The posterior probability that the 2^nd coin was selected is 0.5869.

The posterior probability that the 3^rd coin was selected is 0.3478.

The posterior probability that the 4^th coin was selected is 0.065.

The posterior probability that the 5^th coin was selected is 0.

b. If we continue to toss the same coin until another head is obtained, the probability that exactly three additional tosses will be required is 0.1291.

Step-by-step solution

Given information

Referring to exercise 7, the five coins\(\left( {i = 1,2,3,4,5} \right)\)have different probability of head, stated below:

\({p_1} = 0,{p_2} = \frac{1}{4},{p_3} = \frac{1}{2},{p_4} = \frac{3}{4},{p_5} = 1\)

Here one coin is selected at random from the box and tossed repeatedly until a head appears.

Define the events

The events are:

\({p_i} = \Pr \left( {{H_i}\left| {{C_i}} \right.} \right)\)is probability of head occurring on the i^thcoin is tossed

\({C_i}\)is i^thcoin is tossed

\({H_n}\)is the event that head appears on n^th toss.

Calculate the probability of coin selection

a.

After tossing repeatedly, the first head was obtained on the 4^th toss.

It is required to obtain the following probabilities;

\(\Pr \left( {{C_1}\left| {{H_4}} \right.} \right),\Pr \left( {{C_2}\left| {{H_4}} \right.} \right),\Pr \left( {{C_3}\left| {{H_4}} \right.} \right),\Pr \left( {{C_4}\left| {{H_4}} \right.} \right),\Pr \left( {{C_5}\left| {{H_4}} \right.} \right)\)

In general, it is required to calculate the probability of selecting i-th coin, given that the first head obtained on the 4^th toss. That is\(\Pr \left( {{C_i}\left| {{H_4}} \right.} \right)\)

So,

\(\begin{aligned}{c}{\bf{Pr}}\left( {{{\bf{C}}_{\bf{i}}}\left| {{{\bf{H}}_{\bf{4}}}} \right.} \right){\bf{ = }}\frac{{{\bf{Pr}}\left( {{{\bf{C}}_{\bf{i}}}} \right){\bf{Pr}}\left( {{{\bf{H}}_{\bf{4}}}\left| {{{\bf{C}}_{\bf{i}}}} \right.} \right)}}{{\sum\limits_{{\bf{j = 1}}}^{\bf{5}} {{\bf{Pr}}\left( {{{\bf{C}}_{\bf{j}}}} \right){\bf{Pr}}\left( {{{\bf{H}}_{\bf{4}}}\left| {{{\bf{C}}_{\bf{j}}}} \right.} \right)} }}\\\frac{{{\bf{Pr}}\left( {{{\bf{C}}_{\bf{i}}}} \right){\bf{Pr}}\left( {{{\bf{H}}_{\bf{4}}}\left| {{{\bf{C}}_{\bf{i}}}} \right.} \right)}}{{{\bf{Pr}}\left( {{{\bf{C}}_1}} \right){\bf{Pr}}\left( {{{\bf{H}}_{\bf{4}}}\left| {{{\bf{C}}_1}} \right.} \right) + {\bf{Pr}}\left( {{{\bf{C}}_2}} \right){\bf{Pr}}\left( {{{\bf{H}}_{\bf{4}}}\left| {{{\bf{C}}_2}} \right.} \right) + ... + {\bf{Pr}}\left( {{{\bf{C}}_5}} \right){\bf{Pr}}\left( {{{\bf{H}}_{\bf{4}}}\left| {{{\bf{C}}_5}} \right.} \right)}}\\ = \frac{{\frac{1}{5}{{\left( {1 - {p_i}} \right)}^3}{p_i}}}{{\frac{1}{5}\sum\limits_{j = 1}^5 {{{\left( {1 - {p_j}} \right)}^3}{p_j}} }}\\ = \frac{{256{{\left( {1 - {p_i}} \right)}^3}{p_i}}}{{46}}\end{aligned}\)

On substituting the values of probabilities, the following probabilities are obtained for five different coins:

For\(i = 1\),\(\Pr \left( {{C_1}\left| {{H_4}} \right.} \right) = 0\)

For\(i = 2\),\(\begin{aligned}{c}\Pr \left( {{C_2}\left| {{H_4}} \right.} \right) = \frac{{256 \times \frac{{27}}{{256}}}}{{46}}\\ = 0.58695\end{aligned}\)

For\(i = 3\),\(\Pr \left( {{C_3}\left| {{H_4}} \right.} \right) = 0.3478\)

For\(i = 4\),\(\Pr \left( {{C_4}\left| {{H_4}} \right.} \right) = 0.0652\)

For\(i = 5\),\(\Pr \left( {{C_5}\left| {{H_4}} \right.} \right) = 0\)

Thus, we obtain the probabilities of selecting the i^th coin when the head is obtained on fourth toss.

Calculate the probability of three tosses required

b.

The probability that exactly 3 tosses will be required to obtain second head on i-th coin toss is expressed as,\(\Pr \left( {{C_i}\left| {{H_4}} \right.} \right){\left( {1 - {p_i}} \right)^2}{p_i}\).

So, the probability is,

\(\begin{aligned}{c}{\bf{Pr}}\left( {{\bf{n = 3}}\left| {{{\bf{H}}_{\bf{4}}}} \right.} \right){\bf{ = }}\sum\limits_{{\bf{i = 1}}}^{\bf{5}} {\left[ {{\bf{Pr}}\left( {{{\bf{C}}_{\bf{i}}}\left| {{{\bf{H}}_{\bf{4}}}} \right.} \right){{\left( {{\bf{1 - }}{{\bf{p}}_{\bf{i}}}} \right)}^{\bf{2}}}{{\bf{p}}_{\bf{i}}}} \right]} \\ = \left( \begin{aligned}{c}0 + \left( {0.5869 \times {{\left( {1 - \frac{1}{4}} \right)}^2}\left( {\frac{1}{4}} \right)} \right) + \left( {0.3478 \times {{\left( {1 - \frac{1}{2}} \right)}^2}\left( {\frac{1}{2}} \right)} \right)\\ + \left( {0.0652 \times {{\left( {1 - \frac{3}{4}} \right)}^2}\left( {\frac{3}{4}} \right)} \right) + \left( {0 \times {{\left( {1 - 1} \right)}^2}\left( 1 \right)} \right)\end{aligned} \right)\\ = 0 + \left( {0.5869 \times \frac{9}{{64}}} \right) + \left( {0.3478 \times \frac{1}{8}} \right) + \left( {0.0652 \times \frac{3}{{64}}} \right) + 0\\ = 0.1291\end{aligned}\)

Thus, there is a probability of 0.1291 that exactly 3 additional tosses would be required to obtain the second head.