91Ó°ÊÓ

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.

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.

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.

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.