PB031: Miscellaneous Examples – IV

     Example: 7 Diffculty: Easy      

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

\dfrac{1}{2}\;\left[ {\dfrac{{{{(x + y)}^n} + {{(x - y)}^n}}}{{{{(x + y)}^n}}}} \right]
Solution: 7

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} = \;\mathop {^n{C_r}}\limits_{\scriptstyle{\rm{select}}\;r\;{\rm{things}}\;\hfill\atop  \scriptstyle{\rm{    out}}\;{\rm{of}}\;n\hfill}  \times \mathop {{x^{n - r}}}\limits_{\scriptstyle{\rm{      distribute}}\;{\rm{the}}\hfill\atop  {\scriptstyle{\rm{     remaining}}\;(n - r)\hfill\atop  \scriptstyle{\rm{ things}}\;{\rm{among}}\;{\rm{the}}\;{\rm{boys}}\hfill}}  \times \mathop {{y^r}}\limits_{\scriptstyle{\rm{   distribute}}\;{\rm{the}}\hfill\atop  {\scriptstyle\,\,\,\,r\;{\rm{things}}\;{\rm{among}}\hfill\atop  \scriptstyle{\rm{      the}}\;{\rm{girls}}\hfill}}

What we need to do is simply evaluate {W_0} + {W_2} + {W_4} +\ldots . This can be done as seen in the discussion on Binomial theorem (appendix):

{(x + y)^n} = {\;^n}{C_0}\,{x^n} + {\;^n}{C_1}\,{x^{n - 1}}\,y + {\ldots ^n}{C_r}\,{x^{n - r}}{y^r} +\ldots
{(x - y)^n} = {\;^n}{C_0}\,{x^n} - {\;^n}{C_1}\,{x^{n - 1}}\,y + \ldots  {( - 1)^r}{\;^n}{C_r}\,{x^{n - r}}{y^r} +\ldots

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

\dfrac{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} +\ldots

Thus, the required probability is

\dfrac{{{W_0} + {W_2} + {W_4} +\ldots }}{{{{(x + y)}^n}}}
 = \dfrac{1}{2}\;\left[ {\dfrac{{{{(x + y)}^n} + {{(x - y)}^n}}}{{{{(x + y)}^n}}}} \right]
     Example: 8 Diffculty: Easy      

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 \dfrac{p}{2} otherwise it remains the same. Find the probability that he will qualify.

Solution: 8

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

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 \dfrac{p}{2} = \dfrac{{{p^2}}}{2}(1 - p)
P\left( {{{\bar E}_1}\;{E_2}\;{E_3}} \right) = (1 - p) \times \dfrac{p}{2} \times p = \dfrac{{{p^2}}}{2}(1 - p)
P\left( {{E_1}\;{E_2}\;{E_3}} \right) = p \times p \times p = {p^3}

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

P(E) = 2{p^2} - {p^3}
     Example: 9 Diffculty: Easy      

If 6n balls numbered 0, 1, 2, \ldots , 6n - 1 are placed in a bag and three are drawn at random without replacement, find the probability that the sum of the three numbers an the balls is 6n.

Solution: 9

The important thing to realise here is that we cannot use the integral equation

{x_1} + {x_2} + {x_3} = 6n \ldots(1)

to count the number of solutions here. This is because this equation also counts those solution in which the variables might have repeated values. For example, {x_1} = {x_2} = {x_3} = 2n is a possible solution to (1) , but it is clearly not admissible in the present case.

What we then do is count the number of solutions explicitly, using the lowest number as our ‘anchor.’

Number on ball 1 Numbers on the other two balls Number of solutions
1,6n - 1
2,6n - 2
0 3,6n - 3 3n-1
\vdots
3n-1,3n + 1
2,6n - 3
3,6n - 4
1 \vdots 3n - 2
3n-2,3n + 1
3n-1,3n
3,6n - 5
4,6n - 4
2 \vdots 3n - 4
3n - 2,3n
3n-1, 3n-1
4, 6n - 7
3 5, 6n - 8 3n - 5
\vdots
3n-2, 3n - 1
\begin{array}{l}  \,\,\,\,\, \vdots \\  2n - 1  \end{array} 2n, 2n+1 \begin{array}{l}  \,\,\,\,\, \vdots \\  \,\,\,\,\,1  \end{array}

Go through this table thoroughly. In particular, notice carefully the terms in the last column.

We thus have the total number of favorable solutions as

\left\{ {(3n - 1) + (3n - 2)} \right\} + \left\{ {(3n - 4) + (3n - 5)} \right\} +\ldots \left\{ {(5 + 4)} \right\} + \left\{ {(2 + 1)} \right\}
 = \left( {6n - 3} \right) + \left( {6n - 9} \right) + \ldots 9 + 3 an A.P.
 = 3{n^2}

The total number of ways of choosing 3 balls out of 6n is ^{6n}{C_3}. The required probability p is therefore

p = \dfrac{{3{n^2}}}{{^{6n}{C_3}}}
 = \dfrac{{3n}}{{\left( {6n - 1} \right)\;\left( {6n - 2} \right)}}
0 comments