Answer
Verified
413.1k+ views
Hint: Here a biased coin has a higher probability of heads or tails. A fair coin is a mythical gadget which has probability exactly ½ of showing heads and 1/2of showing tails. First write the set of outcomes so that \[X\]wins, then find out the probability. Similarly, do the same to \[Y\]. Then equate the probabilities of\[X\]wins and \[Y\]wins. So, use this concept to reach the solution of the problem.
Complete step-by-step answer:
\[X\] wins, when the outcome is one of the following set of outcomes:
\[H,TTH,TTTTH,...........\]
Since subsequent tosses are independent, the probability that \[X\] wins is
\[ \Rightarrow P\left( H \right) + P\left( {TTH} \right) + P\left( {TTTTH} \right) + ..................\]
As the probability of showing heads is \[p\], we have
\[
\Rightarrow p + \left( {1 - p} \right)\dfrac{1}{2}p + {\left( {1 - p} \right)^2}\dfrac{1}{{{2^2}}}p + ...................... \\
\Rightarrow p\left[ {1 + \left( {\dfrac{{1 - p}}{2}} \right) + {{\left( {\dfrac{{1 - p}}{2}} \right)}^2} + .............} \right] \\
\]
As the terms are in infinity G.P the sum of the terms are given by \[{S_\infty } = \dfrac{a}{{1 - r}}\] where \[a\] is the first term and \[r\]is the common ratio of the infinity series.
\[
\Rightarrow p\left[ {\dfrac{1}{{1 - \left( {\dfrac{{1 - p}}{2}} \right)}}} \right] \\
\Rightarrow \dfrac{p}{{1 - \dfrac{{1 - p}}{2}}} = \dfrac{{2p}}{{1 + p}} \\
\]
So, the probability of \[X\]wins are \[\dfrac{{2p}}{{1 + p}}....................................\left( 1 \right)\]
Similarly, \[Y\] wins if the outcome is one of the following:
\[TH,TTTH,TTTTTH,....................\]
We know that the sum of probabilities of showing head and showing tail is equal to 1.
So, the probability of showing tile is \[1 - p\], we have
\[
\Rightarrow P\left( H \right) + P\left( {TTH} \right) + P\left( {TTTTTH} \right) + ......................... \\
\Rightarrow \dfrac{{1 - p}}{2} + \dfrac{{{{\left( {1 - p} \right)}^2}}}{4} + \dfrac{{{{\left( {1 - p} \right)}^3}}}{8} + ............................ \\
\Rightarrow \left( {\dfrac{{1 - p}}{2}} \right)\left[ {1 + \dfrac{{1 - p}}{2} + {{\left( {\dfrac{{1 - p}}{2}} \right)}^2} + .........................} \right] \\
\]
As the terms are in infinity G.P the sum of the terms are given by \[{S_\infty } = \dfrac{a}{{1 - r}}\] where \[a\] is the first term and \[r\]is the common ratio of the infinity series.
\[
\Rightarrow \left( {\dfrac{{1 - p}}{2}} \right)\left[ {\dfrac{1}{{1 - \left( {\dfrac{{1 - p}}{2}} \right)}}} \right] \\
\Rightarrow \dfrac{{\dfrac{{1 - p}}{2}}}{{1 - \left( {\dfrac{{1 - p}}{2}} \right)}} = \dfrac{{1 - p}}{{1 + p}} \\
\]
So, the probability of \[Y\]wins are \[\dfrac{{1 - p}}{{1 + p}}.....................................\left( 2 \right)\]
We know that
Probability of \[X\]wins = Probability of \[Y\]wins
\[
\Rightarrow \dfrac{{2p}}{{1 + p}} = \dfrac{{1 - p}}{{1 + p}} \\
\Rightarrow 2p = 1 - p \\
\Rightarrow 2p + p = 1 \\
\Rightarrow 3p = 1 \\
\therefore p = \dfrac{1}{3} \\
\]
Thus, the value of \[p\] is \[\dfrac{1}{3}\].
Note: The terms which are in infinity G.P the sum of the terms are given by \[{S_\infty } = \dfrac{a}{{1 - r}}\] where \[a\] is the first term and \[r\]is the common ratio of the infinity series. The sum of probabilities of showing head and showing tail is equal to one.
Complete step-by-step answer:
\[X\] wins, when the outcome is one of the following set of outcomes:
\[H,TTH,TTTTH,...........\]
Since subsequent tosses are independent, the probability that \[X\] wins is
\[ \Rightarrow P\left( H \right) + P\left( {TTH} \right) + P\left( {TTTTH} \right) + ..................\]
As the probability of showing heads is \[p\], we have
\[
\Rightarrow p + \left( {1 - p} \right)\dfrac{1}{2}p + {\left( {1 - p} \right)^2}\dfrac{1}{{{2^2}}}p + ...................... \\
\Rightarrow p\left[ {1 + \left( {\dfrac{{1 - p}}{2}} \right) + {{\left( {\dfrac{{1 - p}}{2}} \right)}^2} + .............} \right] \\
\]
As the terms are in infinity G.P the sum of the terms are given by \[{S_\infty } = \dfrac{a}{{1 - r}}\] where \[a\] is the first term and \[r\]is the common ratio of the infinity series.
\[
\Rightarrow p\left[ {\dfrac{1}{{1 - \left( {\dfrac{{1 - p}}{2}} \right)}}} \right] \\
\Rightarrow \dfrac{p}{{1 - \dfrac{{1 - p}}{2}}} = \dfrac{{2p}}{{1 + p}} \\
\]
So, the probability of \[X\]wins are \[\dfrac{{2p}}{{1 + p}}....................................\left( 1 \right)\]
Similarly, \[Y\] wins if the outcome is one of the following:
\[TH,TTTH,TTTTTH,....................\]
We know that the sum of probabilities of showing head and showing tail is equal to 1.
So, the probability of showing tile is \[1 - p\], we have
\[
\Rightarrow P\left( H \right) + P\left( {TTH} \right) + P\left( {TTTTTH} \right) + ......................... \\
\Rightarrow \dfrac{{1 - p}}{2} + \dfrac{{{{\left( {1 - p} \right)}^2}}}{4} + \dfrac{{{{\left( {1 - p} \right)}^3}}}{8} + ............................ \\
\Rightarrow \left( {\dfrac{{1 - p}}{2}} \right)\left[ {1 + \dfrac{{1 - p}}{2} + {{\left( {\dfrac{{1 - p}}{2}} \right)}^2} + .........................} \right] \\
\]
As the terms are in infinity G.P the sum of the terms are given by \[{S_\infty } = \dfrac{a}{{1 - r}}\] where \[a\] is the first term and \[r\]is the common ratio of the infinity series.
\[
\Rightarrow \left( {\dfrac{{1 - p}}{2}} \right)\left[ {\dfrac{1}{{1 - \left( {\dfrac{{1 - p}}{2}} \right)}}} \right] \\
\Rightarrow \dfrac{{\dfrac{{1 - p}}{2}}}{{1 - \left( {\dfrac{{1 - p}}{2}} \right)}} = \dfrac{{1 - p}}{{1 + p}} \\
\]
So, the probability of \[Y\]wins are \[\dfrac{{1 - p}}{{1 + p}}.....................................\left( 2 \right)\]
We know that
Probability of \[X\]wins = Probability of \[Y\]wins
\[
\Rightarrow \dfrac{{2p}}{{1 + p}} = \dfrac{{1 - p}}{{1 + p}} \\
\Rightarrow 2p = 1 - p \\
\Rightarrow 2p + p = 1 \\
\Rightarrow 3p = 1 \\
\therefore p = \dfrac{1}{3} \\
\]
Thus, the value of \[p\] is \[\dfrac{1}{3}\].
Note: The terms which are in infinity G.P the sum of the terms are given by \[{S_\infty } = \dfrac{a}{{1 - r}}\] where \[a\] is the first term and \[r\]is the common ratio of the infinity series. The sum of probabilities of showing head and showing tail is equal to one.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are
Assertion The resistivity of a semiconductor increases class 13 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
What is the stopping potential when the metal with class 12 physics JEE_Main
The momentum of a photon is 2 times 10 16gm cmsec Its class 12 physics JEE_Main
Using the following information to help you answer class 12 chemistry CBSE
Trending doubts
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE
Select the correct plural noun from the given singular class 10 english CBSE
What organs are located on the left side of your body class 11 biology CBSE
The sum of three consecutive multiples of 11 is 363 class 7 maths CBSE
What is the z value for a 90 95 and 99 percent confidence class 11 maths CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
How many squares are there in a chess board A 1296 class 11 maths CBSE