1,954 views
3 votes
3 votes
If a fair coin is tossed until the same result turns up in succession (both head or both tail) then find the probability when the number of tosses are even. (Round up to two decimal places)

2 Answers

Best answer
6 votes
6 votes

 here only $1$ coin. 

we tossed several times and observed the result

i.e. first time$==>\left\{H,T \right \}$

Second time $==>\left \{HH,TT,HT,TH \right \}$

third time $==>\left \{HHH,TTH,HTH,THH,HHT,TTT,HTT,THT \right \}$

all are independent

We want both Head or Tail in even number of tosses.

lets take both head case(even number of tosses):-

Even number of tosses means$:2nd$ time,$4th$ time$,6th$ time ,........and so on

$2nd$ $time==>\left \{HH\right \}==>P(HH)=\frac{1}{2}\times\frac{1}{2}=\frac{1}{4}=(\frac{1}{2})^{2}$

$4th$ $time==>\left \{HTHH\right \}==>P(HTHH)=\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}=\frac{1}{16}=(\frac{1}{2})^{4}$

$6th$ $time==>\left \{HTHTHH\right \}==>P(HTHTHH)=\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}=\frac{1}{64}=(\frac{1}{2})^{6}$

$8th$ $time==>\left \{HTHTHTHH\right \}==>P(HTHTHTHH)=(\frac{1}{2})^{8}$

_______________________________________________

                                      $(OR)$

 

lets take both tail case(even number of tosses):-

Even number of tosses means$:2nd$ time,$4th$ time$,6th$ time ,........and so on

$2nd$ $time==>\left \{TT\right \}==>P(TT)=\frac{1}{2}\times\frac{1}{2}=\frac{1}{4}=(\frac{1}{2})^{2}$

$4th$ $time==>\left \{THTT\right \}==>P(THTT)=\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}=\frac{1}{16}=(\frac{1}{2})^{4}$

$6th$ $time==>\left \{THTHTT\right \}==>P(THTHTT)=\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}=\frac{1}{64}=(\frac{1}{2})^{6}$

$8th$ $time==>\left \{THTHTHTT\right \}==>P(THTHTHTT)=(\frac{1}{2})^{8}$

________________________________________________

 

Now even number of tosses until both heads are coming $=(\frac{1}{2})^{2}+(\frac{1}{2})^{4}+(\frac{1}{2})^{6}+(\frac{1}{2})^{8}+........$

This is infinite Geometric Series $S=\frac{a}{1-r}$ When $r<1$ and $S=\infty $ When $r>1$

Here $a=(\frac{1}{2})^{2}$ and $r=(\frac{1}{2})^{2}$

So,$S_{1}=\left [\frac{(\frac{1}{2})^{2}}{1-(\frac{1}{2})^{2}} \right ]$

     $S_{1}=\left [ \frac{\frac{1}{4}}{1-\frac{1}{4}}\right ]$

      $S_{1}=\left [ \frac{\frac{1}{4}}{\frac{3}{4}}\right ]$

    $S_{1}=\left [{ \frac{1}{4} }\times { \frac{4}{3} }\right ]$

     $S_{1}=\frac{1}{3}=0.33$

Similarly for both tails comming in even number of tosses $S_{2}=\frac{1}{3}=0.33$

Now,both heads or tails comming in even number of tosses $S=S_{1}+S_{2}$

                                                                                                   $S=0.33+0.33=0.66$

edited by
4 votes
4 votes
Probability of halting after 2 coin tosses = $\large \frac{1}{2}$, $(HH,TT)$

Probability of halting after 3 coin tosses = $\large \frac{1}{2} \times \frac{1}{2} $, ($HTT,THH)$

Probability of halting after 4 coin tosses = $\large \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}$ $(HTHH,THTT)$

So from this we can see that,

Probability of halting after n coin tosses = $\large (\frac{1}{2})^{(n-1)}$

Sum of all halting even coin tosses = $\large \frac{1}{2} + (\frac{1}{2})^3 + (\frac{1}{2})^5 + (\frac{1}{2})^7 + .... $

this a GP with a = $(\frac{1}{2})$ and r = $(\frac{1}{2})^2$

Sum of an infinite GP = $\large \frac{a}{1-r}$

 $ = \Large \frac{\frac{1}{2}}{1- \frac{1}{2}^2}$

 $ = \Large \frac{0.5}{0.75} $$ = 0.66$

Related questions

1 votes
1 votes
1 answer
1
sim1234 asked Dec 26, 2018
683 views
Assuming that a pointer take 4 bytes and the size of an integer is 2 bytes. What is the size of the *a in declaration: int (*a) [10] ? 4802040
0 votes
0 votes
0 answers
2
sim1234 asked Dec 21, 2018
690 views
Suppose R is a table with 4 attributes A,B, C and D, and S be a table with 3 attributes B, E, F. Now given a particular instance of R and S.S is empty, having no tuples. ...
0 votes
0 votes
0 answers
3
sim1234 asked Dec 17, 2018
396 views
Database that are designed and managed specifically to meet information needs are calledDatabase Management systemData WarehousesTransaction databasesProduction databases...
0 votes
0 votes
1 answer
4
sim1234 asked Dec 15, 2018
679 views
Consider CSMA/CD protocol, after 4th collision, what is the probability that node choose k=4.