# Probability Set 2

**Example – 5**

Sixteen players *S*_{1}, *S*_{2}, *S*_{3},......*, S*_{16} 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 *S*_{1} is among the eight winners.

**(b)** Find the probability that exactly one of *S*_{1 }and *S*_{2} is among the eight winners.

**Solution:** **(a)*** S*_{1} can be paired with any of the remaining 15 players, say *S j*. Now,

*S*

_{1}' s game with

*S*is independent of the other seven games. This means that the winner of the

**j***S*

_{1 }–

*S*

_{j}_{ }game can be either

*S*

_{1}or

*S*

_{j}_{ }with equal probability. The probability of

*S*

_{1}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 *S*_{1} and *S*_{2} should be among the eight winners. Two cases arise:

Scenario |
Probability |

Case - 1 S_{1 }paired with S_{2} ; one wins |
\[\frac{1}{15}\qquad\left\{ \begin{align} & {{S}_{1}}\,\text{can be paried in 15 } \\ & \text{equally likely ways, } \\ & \text{of which 1 is favorable} \\ \end{align} \right\}\] |

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

Thus, the probability *p* that exactly one of *S*_{1} and *S*_{2} 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 *P _{n}* 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 *n*th ‘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 (*n* – *r*) are given to the boys. The number of ways *W _{r}*

_{ }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}

\end{smallmatrix}}{\mathop{{{y}^{r}}}}\,\]

What we need to do is simply evaluate *W*_{0} + *W*_{2 }+ *W*_{4} +.............. . 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 *E _{i}* the event that the student passes the

*i*

^{th}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

\[P(E)=2{{p}^{2}}-{{p}^{3}}\]

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