Latest Elitmus Aptitude Question SOLUTION: 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

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(#M40018474) ELITMUS Numerical Ability question Ratio and Proportion Keep an EYE

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)

Asked In Elitmus Sumit Rawal (11 years ago)

Unsolved Read Solution (16)

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Sneha Agarwal  (10 years ago)

answer is 1:2

Is this solution Helpfull? Yes (23) | No (5)  

Saiyam Behl  (10 years ago)

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

Is this solution Helpfull? Yes (12) | No (3)  

Sumit Rawal  (11 years ago)

@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"

Is this solution Helpfull? Yes (9) | No (3)  

Gaurav Tiwari  (10 years ago)

odds in favor = prob of winning / prob. of loosing
=(m/n+m)/(n/n+m)=3/2
m/n=3/2
ans : 3:2

Is this solution Helpfull? Yes (7) | No (1)  

Shivani Thakur  (8 years ago)

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

Is this solution Helpfull? Yes (6) | No   

Gourango  (10 years ago)

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:-)

Is this solution Helpfull? Yes (5) | No   

DIKSHA RANGNANI  (10 years ago)

odds in favour of his winning the game is 3/2= what does this means ?
and please tell me the answer ..

Is this solution Helpfull? Yes (2) | No   

HARISHANKAR YADAV  (9 years ago)

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

Is this solution Helpfull? Yes (1) | No   

santosh  (8 years ago)

what is the right answer ???1:2 or 3:2

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naina  (11 years ago)

@sumit are u sure this was asked in june 23 2013??

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manisha nayak  (10 years ago)

@sneha Can you please explain how you got the answer

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Pratyush  (10 years ago)

[m/(m+n)]/[nm/(m+n)^2]
=(m+n)/n
=m/n +1=3/2
=>m/n=1/2 Ans=1:2

Is this solution Helpfull? Yes | No (1)  

prasanthi  (9 years ago)

ODDS IN FAVOUR MEANS P(ABHISHER/AISHWARYA)=P(ABHISHEK)*P(AISH)/P(AISH)=M/M+N*N/M+N=N/M+N

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santosh  (8 years ago)

what is the right answer ? b or d

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KUMARA S K  (7 years ago)

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

Is this solution Helpfull? Yes | No (1)  

Digesh Singh Bisht  (1 year ago)

@Kumar S K .. donot copy paste others answer.

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