IIT JEE , AIEEE Forum - Mathematics , Algebra

bsgdabest

A fair coin is tossed 13 times. What is the probabilty that two heads donot occur consecutively? ( correct solution: 20 pts) State ur answer with proof

sboosy

This sum can be solved by obtaining a sequence ..from which it is easy to generalize

Consider a fair die tossed twice

u could get

HT ,TT ,TH,HH

the answer will be 3/4

Consider a fair die tossed thrice

u could get

HHH,HHT,HTH,HTT,THH,TTH,THT,TTT

answer is 5/8

similarly consider it tossed 4 times

the answer is 9/16

Did u get the sequence?

the numerator is always the (denominator/2+1)

so if u toss 13 times what is denominator?

2¹³

so answer is

[(2¹³/2)+1]/2¹³

= (2¹²+1)/2¹³

akhil_o

the ways of getting the throws such that no heads occur simultaneously:

13 T
12 T 1 H
11 T 2 H
10 T 3 H
.
..
.

for arranging we have to place t tails and place h heads in the t+1 gaps created
so for any case
^t+1C_h
so we can see it as
¹⁴C₀+¹³C₁+¹²C₂+......⁷C₇

total no of cases is 2¹³
so probablity is
(¹⁴C₀+¹³C₁+¹²C₂+......⁷C₇)/2¹³

akhil_o

@sboosy...
i am really sorry i couldnt get 9 cases in the 4 throws?

TTTT- 1 case
THHT-3 cases(THTH HTHT HTTH)
TTTH-4 cases
that makes 8 cases not 9

similarly for 5 coins

TTTTT- 1 case
TTTTH- 5 cases
TTTHH-6 cases (10 ways of arranging- 4 unfavorable)
TTHHH- only 1 (HTHTH)
that makes 13 only and not 17 as u require it to be
3/4 is right... but after that ur series fails to hold
so i still believe my answer is right...coming out to
610/2^13

PS plz correct me if i am wrong!might have missed sumthing here!

general solution:
for n tosses:
summation(from i=0 to i=(n+1)/2){^n+1-iC_i}for n =2k+1
summation(from i=0 to i=(n)/2){^n+1-iC_i} for n=2k

akhil_o

Anyone has the answer?

nadeemoidu

I guess the answer given by akhil is right.

Here is a method which is not included in JEE but is very easy to understand. Both the methods give the same answer.

Let h(x) denote the no. of possible combinations with H at the end and t(x) denote the no. of possible combination with T at the end.

So the final answer will be h(13) + t(13)

Now, h(x) = t(x-1)    Because the only way to get a head at the end is to put H at the end of series ending with T.

Similarly t(x) = h(x-1 ) + t(x-1) because u can put a T at the end of any series.

so we can manually calculate the values upto 13

x    1   2 3 4 5 6     7    8
h(x) 1 1 2 3 5 8    13 21
t(x) 1 2 3 5 8 13 21 34

it is easy to see that this is the fibonacci series. h(13 ) + t ( 13 ) will be 610 .

akhil_o

brilliant nadeemoidu...we missed fibonacci!

bsgdabest

ans. is 610/2^13

bsgdabest

didnt understand dude.. what is h(x) ? is it no. of outcomes where H appears at the xth position in 13 tosses?

nadeemoidu

h(x) is the no. of ways in which H and T appear in 'x' throws such that the last throw is H and no two Hs are consecutive.

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