Abhishek and Aishwarya picked up a ball at random from a bag of M red colour and N yellow coloured

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
11 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
11 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
9 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
10 years agoHelpfull: Yes(1) No(0)
what is the right answer ???1:2 or 3:2
9 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
11 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
11 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
10 years agoHelpfull: Yes(0) No(0)
what is the right answer ? b or d
9 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
8 years agoHelpfull: Yes(0) No(1)
@Kumar S K .. donot copy paste others answer.
2 years agoHelpfull: Yes(0) No(0)

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