Example – 5

Sixteen players S1, S2, S3,......, S16  play in a tournament. They are divided into eight pairs at random. From each pair a winner is decided on the basis of a game played between the two players of the pair. Assume that all the pairs are of equal strength.

(a) Find the probability that the players S1 is among the eight winners.

(b) Find the probability that exactly one of S1 and S2 is among the eight winners.

Solution: (a) S1 can be paired with any of the remaining 15 players, say Sj. Now, S1' s game with Sj is independent of the other seven games. This means that the winner of the S1 Sj game can be either S1 or Sj with equal probability. The probability of S1 winning (and being among the eight winners) is therefore simply \(\frac{1}{2}\). In fact, this is the probability for each of the sixteen players.

(b) Now we want that only one of S1 and S2 should be among the eight winners. Two cases arise:

Scenario Probability
Case - 1     S1 paired with S2 ; one     wins \[\frac{1}{15}\qquad\left\{ \begin{align} & {{S}_{1}}\,\text{can be paried in 15 } \\ & \text{equally likely ways, } \\ & \text{of which 1 is favorable} \\ \end{align} \right\}\]

Case - 2    S1 not paired with S

                 Now, either

                S1 wins, S2 loses


                 S1 loses, S2 wins

\[\begin{align}& \frac{14}{15}\times \frac{1}{2}\times \frac{1}{2}=\frac{7}{30} \\  &  \\ 
 & \frac{14}{15}\times \frac{1}{2}\times \frac{1}{2}=\frac{7}{30} \\ \end{align}\]

Thus, the probability p that exactly one of S1 and S2 is among the eight winners is

\[\begin{align}& p=\frac{1}{15}+\frac{7}{30}+\frac{7}{30} \\ &\; =\frac{8}{15} \\ \end{align}\]

Example – 6

A player tosses a coin and scores 1 point for a head and 2 points for a tail. He plays in until his score reaches or passes n. If Pn denotes the probability of getting a score of exactly n, show that

\[{{P}_{n}}=\frac{1}{3}\ \left[ 2+{{(-1)}^{n}}\frac{1}{{{2}^{n}}} \right]\]

Solution: First of all, note that    \(\begin{align}{{P}_{1}}=\frac{1}{2}\,\,and\,\,{{P}_{2}}=\frac{3}{4}\,\,\left( why? \right)\end{align}\)

Both these values satisfy the general relation which we have to prove to be true.

Now, the form of the problem suggests that a recursion can be easily set up. To write the recursion, note that a score of exactly n can be reached in two mutually exclusive ways (the corresponding probabilities are also mentioned):

(1) A score of exactly (n – 1), then a Head : \(\begin{align}{{P}_{n-1}}\times \frac{1}{2}\end{align}\)

(2) A score of exactly (n – 2), then a Tail  :  \(\begin{align}{{P}_{n-2}}\times \frac{1}{2}\end{align}\)

Thus, we simply have

\[{{P}_{n}}=\frac{1}{2}\left( {{P}_{n-1}}+{{P}_{n-2}} \right)\]

This recursion relation, upon some rearrangement, gives:

\[\begin{align} &\qquad\quad {{P}_{n}}+\frac{1}{2}\ {{P}_{n-1}}={{P}_{n-1}}+\frac{1}{2}\ {{P}_{n-2}} \\ &\qquad\qquad\quad\;\qquad\quad={{P}_{n-2}}+\frac{1}{2}\ {{P}_{n-3}} \\ &\qquad\qquad\quad\qquad\qquad \qquad\quad \vdots  \\  & \qquad\qquad\quad\;\qquad\quad={{P}_{2}}+\frac{1}{2}\,{{P}_{1}} \\  & \qquad\qquad\quad\;\qquad\quad=1 \\  & \Rightarrow \quad {{P}_{n}}+\frac{1}{2}\ {{P}_{n-1}}=1 \\  & \Rightarrow \quad {{P}_{n}}-\frac{2}{3}=\frac{1}{3}-\frac{1}{2}\ {{P}_{n-1}}=\frac{-1}{2}\,\left( {{P}_{n-1}}-\frac{2}{3} \right) \\ \end{align}\]

In the last step, why we did the indicated manipulation is very important to understand. What we have done is express \(\begin{align}\left( {{P}_{n}}-\frac{2}{3} \right)\end{align}\) in terms of \(\begin{align}\left( {{P}_{n-1}}-\frac{2}{3} \right)\end{align}\) , which effectively means expressing the nth ‘term’ in terms of the (n – 1)th ‘term’. Why does this help us? It is useful because we can now apply it repeatedly till we exhaust the possibility of repetition:

\[\begin{align}\qquad\qquad\;& {{P}_{n}}-\frac{2}{3}=\frac{-1}{2}\left( {{P}_{n-1}}-\frac{2}{3} \right) \\ & \qquad\quad\;\; ={{\left( \frac{-1}{2} \right)}^{2}}\ \left( {{P}_{n-2}}-\frac{2}{3} \right) \\ 
 & \qquad\quad\;\; ={{\left( \frac{-1}{2} \right)}^{3}}\ \left( {{P}_{n-3}}-\frac{2}{3} \right) \\  & \qquad\qquad \qquad\qquad \vdots  \\  & \qquad\quad\;\; ={{\left( \frac{-1}{2} \right)}^{n-1}}\ \left( {{P}_{1}}-\frac{2}{3} \right) \\ & \qquad\quad\;\; =\frac{-1}{6}\cdot {{\left( \frac{-1}{2} \right)}^{n-1}} \\ &\qquad\quad\;\; =\frac{{{(-1)}^{n}}}{{{3.2}^{n}}} \\ & \Rightarrow \quad {{P}_{n}}=\frac{1}{3}\ \left[ 2+{{(-1)}^{n}}\ \frac{1}{{{2}^{n}}} \right] \\ \end{align}\]

which proves the assertion.

You are urged to reread the solution until you fully understand how the recursion was setup and solved.

Example – 7

If n different things are distributed among x boys and y girls, show that the probability that the number of things received by the girls is even is

\[\frac{1}{2}\ \left[ \frac{{{(x+y)}^{n}}+{{(x-y)}^{n}}}{{{(x+y)}^{n}}} \right]\]

Solution: One individual thing can be assigned in a total of (x + y) ways, so that n different things can be distributed in (x + y)n ways.

Now, suppose r things are given to girls and the rest (nr) are given to the boys. The number of ways Wr of doing this is

\[{{W}_{r}}=\ \underset{\begin{smallmatrix} 
 \text{select}\ r\ \text{things}\  \\ 
 \text{    out}\ \text{of}\ n 
\end{smallmatrix}}{\mathop{^{n}{{C}_{r}}}}\,\times \underset{\begin{smallmatrix} 
 \text{      distribute}\ \text{the} \\ 
 \text{     remaining}\ (n-r) 
 \text{ things}\ \text{among}\ \text{the}\ \text{boys} 
\end{smallmatrix}}{\mathop{{{x}^{n-r}}}}\,\times \underset{\begin{smallmatrix} 
 \text{   distribute}\ \text{the} \\ 
 \,\,\,\,r\ \text{things}\ \text{among} 
 \text{      the}\ \text{girls} 

What we need to do is simply evaluate W0 + W2 + W4 +.............. . This can be done as seen in the discussion on Binomial theorem (appendix):

\[\begin{align}& {{(x+y)}^{n}}={{\ }^{n}}{{C}_{0}}\,{{x}^{n}}+{{\ }^{n}}{{C}_{1}}\,{{x}^{n-1}}\,y+{{.....}^{n}}{{C}_{r}}\,{{x}^{n-r}}{{y}^{r}}+.... \\ & {{(x-y)}^{n}}={{\ }^{n}}{{C}_{0}}\,{{x}^{n}}-{{\ }^{n}}{{C}_{1}}\,{{x}^{n-1}}\,y+.....{{(-1)}^{r}}{{\ }^{n}}{{C}_{r}}\,{{x}^{n-r}}{{y}^{r}}+.... \\ \end{align}\]

Adding the two, we are left with only the even terms:

\[\frac{1}{2}\ \left\{ {{(x+y)}^{n}}+{{(x-y)}^{n}} \right\}={{\ }^{n}}{{C}_{0}}{{x}^{n}}+{{\ }^{n}}{{C}_{2}}{{x}^{n-2}}{{y}^{2}}+{{\ }^{n}}{{C}_{4}}{{x}^{n-4}}{{y}^{4}}+.....\]

Thus, the required probability is

\[\begin{align} & \frac{{{W}_{0}}+{{W}_{2}}+{{W}_{4}}+....}{{{(x+y)}^{n}}} \\ & =\frac{1}{2}\ \left[ \frac{{{(x+y)}^{n}}+{{(x-y)}^{n}}}{{{(x+y)}^{n}}} \right] \\ \end{align}\]

Example – 8

For a student to qualify he must pass at least two out of three examinations. The probability that he will pass the first examination is p. If he fails in one of the examination, then the probability of his passing in the next examination is  \(\frac{p}{2}\) otherwise it remains the same. Find the probability that he will qualify.

Solution: Let us denote by Ei the event that the student passes the ith examination and by E the event that he qualifies. Thus, E can happen in four possible mutually exclusive ways (sequences).

\[{{E}_{1}}\ {{E}_{2}}\ {{\bar{E}}_{3}},\ \ {{E}_{1}}\ {{\bar{E}}_{2}}\ {{E}_{3}},\ \ {{\bar{E}}_{1}}\ {{E}_{2}}\ {{E}_{3}},\ \ {{E}_{1}}\ {{E}_{2}}\ {{E}_{3}}\]

We have

\[\begin{align}& P\left( {{E}_{1}}\ {{E}_{2}}\ {{{\bar{E}}}_{3}} \right)=p\times p\times (1-p)={{p}^{2}}\left( 1-p \right) \\ & P\left( {{E}_{1}}\ {{{\bar{E}}}_{2}}\ {{E}_{3}} \right)=p\times (1-p)\times \frac{p}{2}=\frac{{{p}^{2}}}{2}(1-p) \\ & P\left( {{{\bar{E}}}_{1}}\ {{E}_{2}}\ {{E}_{3}} \right)=(1-p)\times \frac{p}{2}\times p=\frac{{{p}^{2}}}{2}(1-p) \\ & P\left( {{E}_{1}}\ {{E}_{2}}\ {{E}_{3}} \right)=p\times p\times p={{p}^{3}} \\ \end{align}\]

Thus, P(E) is simply obtained by adding the four probabilities which gives


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