1 Star2 Stars3 Stars4 Stars5 Stars (1 votes, average: 5.00 out of 5)
Loading ... Loading ...
Print Print

GATE-2012 ECE Q36 (math)

by Krishna Sankar on October 24, 2012

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

(C) 2/3

(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,

Given that the coin is fair,

.

The total probability is,

Using Taylor series,

.

Substituting,

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] Wiki entry on Taylor series http://en.wikipedia.org/wiki/Taylor_series

 

D id you like this article? Make sure that you do not miss a new article by subscribing to RSS feed OR subscribing to e-mail newsletter. Note: Subscribing via e-mail entitles you to download the free e-Book on BER of BPSK/QPSK/16QAM/16PSK in AWGN.

{ 0 comments… add one now }

Leave a Comment

Previous post:

Next post: