Abhishek and Aishwarya picked up a ball at random from a bag of M red colour and N yellow coloured balls,one after the another, replacing the ball every time till one of them gets a red ball.The first one to get a red ball is declared as a winner.IF Abhishek wins the game and odds in favour of his winning the game is 3/2 find the ratio of M/N
A)1:1
B)1:2
C)2:3
D)3:2
(Asked in ELitmus held on June 23,2013)

• answer is 1:2
  
• 11 years agoHelpfull: Yes(23) No(5)
• odds in favour of an event = (prob. that an event will occur)/(prob. that an event will not occur)
  M/(M+N)=3/5 or Lose(Yellow) N/(M+N)=2/5
  therefore ratio m:n =3/2
• 10 years agoHelpfull: Yes(12) No(3)
• @naina a similar question was asked on 23rd june 2013 when i have taken the exam but 1 thing for sure and for every aspirant "questions do repeat in elitmus"
• 11 years agoHelpfull: Yes(9) No(3)
• odds in favor = prob of winning / prob. of loosing
  =(m/n+m)/(n/n+m)=3/2
  m/n=3/2
  ans : 3:2
• 10 years agoHelpfull: Yes(7) No(1)
• P(abhi) = chances of abhi winning the game is 3/5
  chances of abhi getting red ball is M/(M+N)
  abhi begins the game and wins: so it goes like [(abhi get the red ball) OR (abhi doesn't get red AND aish doesn't get red AND abhi gets the red) OR ........]
  so it can be expressed like
  abhi + (not abhi)(not aish) abhi + (not abhi)(not aish)(not abhi)(not aish) abhi + (not abhi) (not aish) (not abhi)(not aish)(not abhi)(not aish) abhi + .....
  M/(M+N) + (N/(M+N))^2 M/(M+N) + (N/(M+N))^4 M/(M+N) + (N/(M+N))^6 M/(M+N) +............
  =>M/(M+N) [1 + (N/(M+N))^2 + (N/(M+N))^4 + (N/(M+N))^6 +.......]
  
  
  The expression 1 + x^2 + x^4 + .... is an infinite geometric series with common ratio r = x^2 and first term a = 1 . The infinite sum is s = a / (1 - r) = 1 / (1 - x^2)
  => so sum of this series is 1/(1-(N/(M+N))^2)
  
  => M/(M+N) * (M+N)^2 / ((M+N)^2 - N^2)
  =>(M*(M+N))/((M+N)^2-N^2)
  =>(M*(M+N))/(M(M+2N))
  
  
  Now according to the question
  =>(M+N)/(M+2N) = 3/5
  =>5M+5N=3M+6N
  =>M:N=1:2
  
  
  
  
  
• 8 years agoHelpfull: Yes(6) No(0)
• avishek:aishwarya=m/(m+n) :n/(m+n) *m/(m+n)
  =>(m+n) :n
  probability of avishek winning p(E) =(m+n) /(m+n+n)
  p(E') =1-p(E)
  odds in favour =p(E)/p(E')
  
  =(m+n) /n
  =m/n+1=3/2
  =) m/n=1:2
  correct answer:-)
• 10 years agoHelpfull: Yes(5) No(0)
• odds in favour of his winning the game is 3/2= what does this means ?
  and please tell me the answer ..
• 11 years agoHelpfull: Yes(2) No(0)
• prob of win when get red boll=M/M+N
  now P(E)=(M/M+N)
  P(E)'=(1-M/M+N)=N/M+N
  odds in favour=P(E)/P(E)=(M/M+N)/(N/M+N)
  =M/N
  3/2
  
• 9 years agoHelpfull: Yes(1) No(0)
• what is the right answer ???1:2 or 3:2
• 8 years agoHelpfull: Yes(1) No(0)
• @sumit are u sure this was asked in june 23 2013??
• 11 years agoHelpfull: Yes(0) No(0)
• @sneha Can you please explain how you got the answer
• 10 years agoHelpfull: Yes(0) No(0)
• [m/(m+n)]/[nm/(m+n)^2]
  =(m+n)/n
  =m/n +1=3/2
  =>m/n=1/2 Ans=1:2
• 10 years agoHelpfull: Yes(0) No(1)
• ODDS IN FAVOUR MEANS P(ABHISHER/AISHWARYA)=P(ABHISHEK)*P(AISH)/P(AISH)=M/M+N*N/M+N=N/M+N
  
• 9 years agoHelpfull: Yes(0) No(0)
• what is the right answer ? b or d
• 8 years agoHelpfull: Yes(0) No(0)
• P(abhishek) = chances of abhishek wins is
  chances of abhishek getting red ball
  abhishek begins the game and wins: so it goes like [(abhishek get the red ball) OR (abhishek doesn't get red AND aishwarya doesn't get red AND abhishek gets the red) OR ........]
  so it can be expressed like
  abhi + abhi aish abhi + abhi aish abhi aish abhi + abhi aish abhi aish abhi aish abhi + .....
  M/(M+N) + (N/(M+N))^2 M/(M+N) + (N/(M+N))^4 M/(M+N) + (N/(M+N))^6 M/(M+N) +............
  =>M/(M+N) [1 + (N/(M+N))^2 + (N/(M+N))^4 + (N/(M+N))^6 +.......]
  The expression 1 + x^2 + x^4 + .... is an infinite geometric series with common ratio r = x^2 and first term a = 1 . The infinite sum is s = a / (1 - r) = 1 / (1 - x^2)
  => so sum of this series is 1/(1-(N/(M+N))^2)
  => M/(M+N) * (M+N)^2 / ((M+N)^2 - N^2)
  =>(M*(M+N))/((M+N)^2-N^2)
  =>(M*(M+N))/(M(M+2N))
  Now according to the question
  =>(M+N)/(M+2N) = 3/5
  =>5M+5N=3M+6N
  =>2M=N
  =>M/N=1/2
  =>M:N = 1:2
  
  
  
  
• 7 years agoHelpfull: Yes(0) No(1)
• @Kumar S K .. donot copy paste others answer.
• 1 year agoHelpfull: Yes(0) No(0)