Sums on Probability::::::::::

 

1.Find the probability that in 5 tossings a coin turns up head at least 3 times in succession.

2.From the no.s 1,2,.....................,(2n+1) 3 are selected at random.What is the prob. that this no.s are in A.P.

first problem

5 coins tossed

total no.of ways  2*2*2*2*2=2^5=32

0 heads    the no.of ways   =5!\5!=1(TTTTT)

1  head     the no.of ways    = 5!\4!=5( like   HTTTT=5!\4!)

2 heads    the no. of ways   =5!\2!*3!=10(like HHTTT=5!\2!*3!)

3 heads                                   =5!\3!*2!=10(HHHTT)

4 heads                                    =5!\4!=5(HHHHT)

5 heads                                    =5!\5!=1(HHHHH)

THE PROBABILITY IS      (at least 3 heads  10+5+1=16)=>  16\32=0.5.

if any wrong sorry.

Sorry,the answer is wrong.Right answer is 1/4.By the way I have already done this sum so try to crack the other one please

second  problem

let m=2n+1.

now for difference(d)=1.

choices are (1,2,3)(2,3,4)...................................(2n-1,2n,2n+1).

ie the no.of choices are=2n-1 

now for d=2

choices are  (1,3,5),(2,4,6),.....................................(2n-3,2n-1,2n+1)

the choices are       =2n-3.

for d =3  the choices are     =2n-5.

for d=4 the choices are      =2n-7

similarly   

for d=n   the choices are (1,n+1,2n+1)=   1 choice

now sum of all choices are =1+3+5+7+..................+2n-3+2n-1

                                                   =n^2

the probability    =no.of possible ways  \  totalways  

=>n^2\(2n+1)C3