Question

A fair coin is tossed 10 times. Find the probability that two heads do not occur consecutively

Answer 1

Answer:

Probability of getting two heads =

no. of heads /no. of tossing

2/10=1/5.ans

Answer 2

HELLO ,

Possibilities are (dont mind the number of terms):

HTTTTTTH, HTHTHTHTHTHTHT.

But except for those,

let y(n) be the number of sequences that start with T

T _, there are two options, T and H so,

y(n)=y(n−1)+x(n−1)=y(n−1)+y(n−2)

Let x(n) be the number of sequences that start with H,

H _ the next option is only T hence,

x(n)=y(n−1)

The total number of sequences F(n) is:

F(n)=y(n)+x(n)=2y(n−1)+y(n−2)

We are after F(10),

F(10)=2y(9)+y(8)

F(3)=2y(2)+y(1)=1+4=5

F(4)=2y(3)+y(2)=6+2=8

But I'm not quite sure where this will take me though.

BUT I HOPE THIS ANSWER MAY HELP YOU......!