IIT JEE , AIEEE Forum - Mathematics , Algebra

m.viddya

Q.Two persons and A and B toss 2 coins one after the other.The person who throws one head and one tail is the winner .If A starts the game then the probability that B wins the game is?

tango_goat

for B to win the game A should also lose .....probability of A and B winning is 1/2.....so B 's probability of winning the game is 1/4
i hope its right ....
please nudge me if i'm wrong......

rockey

b to win a have to lose.i.e he should get both heads or tails.its probability is 1/4+1/4=1/2.
probability of B winning =1/2
i hope it is correct.
please vote for me

m.viddya

well the answer is given as 1\3...can someone help me out??

shubham_cool

Let probability that a wins be p = 1/2.............[ Sample space=HH,HT,TH,TT

A=HT,TH

B=HTTH]

probability that b wins be q = 1/2

therefore probability that A wins is = p + p'q'p + p'q'p'q'p +............

= p (1 + p'q' + (p'q')^2 + (p'q')^3 +.............)

this forms a gometric progresion to infinity in p'q'

Therefore,

Probability that A wins = p( 1/(1-(p'q')))

= 2/3..................[ p'=1/2 ; q' =1/2 ; p= 1/2]

So, probability that B wins = 1-(2/3)

= 1/3

I DONOT HOPE..........BUT I KNOW ITS WRITE.........

Thank you

pannaguma

simple.
first write sample space
HH
HT
TH
TT

thus in a try probability of
A winning pA = 1/2
A losing p"A = 1/2

similarly for B

now B can win the first time he 'gets' the chance, "or" second time he gets the chance ...............

thus p"A*pB + p"A*p"B*p"A*pB + ........... infinte Gp with
a=1/4
r=1/4

thus Sn= a/1-r = 1/4 / 3/4 = 1/3

plz rate.

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