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GATE-2012 ECE Q36 (math)

Posted By Krishna Sankar On October 24, 2012 @ 3:19 pm In GATE | No Comments

Question 36 on math from GATE (Graduate Aptitude Test in Engineering) 2012 Electronics and Communication Engineering paper.

Q36. A fair coin is tossed till a head appears for the first time. The probability that the number of required tosses is odd, is

(A) 1/3

(B) 1/2

(D) 3/4

Solution

Let us start by finding the sample space. The possible sample space with odd number of tosses till a head appears for the first time is,

$\begin{array}S=\{H,&TTH,& TTTTH,& TTTTTTH,& ...&\}\end{array}$

Given that the coin is fair,

$p(H)=p(T)=\frac{1}{2}$ .

The total probability is,

$\begin{array}{lll}p(S)&=&\frac{1}{2}+\frac{1}{2^3}+\frac{1}{2^5}+\frac{1}{2^7}+...\\&=&\frac{1}{2}$1+\frac{1}{2^2}+\frac{1}{2^4}+\frac{1}{2^6}+...$\\&=&\frac{1}{2}$1+\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+...$\end{array}$

Using Taylor series ^[1],

$\begin{array}\frac{1}{1-x}&=&\sum_{n=0}^{\infty}x^n,&\mbox{for } |x| < 1\end{array}$ .

Substituting,

$\begin{array}{lll}p(S)&=&\frac{1}{2}*\frac{1}{$1-\frac{1}{4}$}\\&=&\frac{2}{3}\end{array}$

Matlab/Octave example

Felt it would be nice to get similar results using a simple simulation model

% Script to find the probability that odd number of tosses
% are required to get a head for the first time

clear all; close all;
N = 5*10^5;

% HEAD = 0, TAIL = 1;
% definfing the pattern corresponding to odd tosses 
% till the first head - limiting to max of 21 tosses
% converted to integer (for easy comparison)
pattern_v = [ ...
   sum(2.^0         .*  [0 ]);                      ... % H
   sum(2.^(2:-1:0)  .*  [1 1 0 ]);                  ... % TTH
   sum(2.^(4:-1:0)  .*  [1 1 1 1 0 ]);              ... % TTTTH 
   sum(2.^(6:-1:0)  .*  [1 1 1 1 1 1 0 ]);          ... % TTTTTTH 
   sum(2.^(8:-1:0)  .*  [1 1 1 1 1 1 1 1 0 ]);      ... % TTTTTTTTH ...
   sum(2.^(10:-1:0) .*  [1 1 1 1 1 1 1 1 1 1 0 ]);  ...  
   sum(2.^(12:-1:0) .*  [1 1 1 1 1 1 1 1 1 1 1 1 0 ]); ...
   sum(2.^(14:-1:0) .*  [1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 ]); ... 
   sum(2.^(16:-1:0) .*  [1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 ]); ... 
   sum(2.^(18:-1:0) .*  [1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 ]); ...
   sum(2.^(20:-1:0) .*  [1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 ]); ...
  ];

% finding the probability for each event in the  sample space
event_v = [1:2:21];
for ii=1:length(event_v)
   kk           = event_v(ii);
   x            = rand(N,kk)>0.5;
   xVal         = sum((ones(N,1)*[2.^([kk-1:-1:0])]).*x,2);
   matchCnt(ii) = size(find(xVal==pattern_v(ii)),1);
end
totalProb = sum(matchCnt./N)

Figure : Probability of odd number of tosses till first head

Based on the above, the right choice is (C) 2/3.

References

[1] GATE Examination Question Papers [Previous Years] from Indian Institute of Technology, Madras http://gate.iitm.ac.in/gateqps/2012/ec.pdf ^[2]

[2] Wiki entry on Taylor series http://en.wikipedia.org/wiki/Taylor_series ^[1]

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[1] Taylor series: http://en.wikipedia.org/wiki/Taylor_series

[2] http://gate.iitm.ac.in/gateqps/2012/ec.pdf: http://gate.iitm.ac.in/gateqps/2012/ec.pdf

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Q36. A fair coin is tossed till a head appears for the first time. The probability that the number of required tosses is odd, is

Solution

References

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