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

• Placement Papers
• /
• Elitmus
• /
• Numerical Ability
• /
• Probability

Elitmus Question Papers:

(#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)

Is this Puzzle helpful?   (1)   (1) Submit Your Solution Probability

Read Solution (16) :

Sneha Agarwal  (11 years ago)

answer is 1:2

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

Saiyam Behl  (11 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  (11 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  (9 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  (11 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  (10 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  (9 years ago)

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

Is this solution Helpfull? Yes (1) | No   

naina  (11 years ago)

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

Is this solution Helpfull? Yes | No   

manisha nayak  (11 years ago)

@sneha Can you please explain how you got the answer

Is this solution Helpfull? Yes | No   

Pratyush  (11 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  (10 years ago)

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

Is this solution Helpfull? Yes | No   

santosh  (9 years ago)

what is the right answer ? b or d

Is this solution Helpfull? Yes | No   

KUMARA S K  (8 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  (2 years ago)

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

Is this solution Helpfull? Yes | No   

Please Login to Read Solution.