Probability question

[badboys]

Originally Posted by sumit82

Can someone help me with this, I can;t get where exactly I am going wrong.....


Q. Kurt, a painter, has 9 jars of paint 4 of which are yellow, 2 are red and the remaining jars are brown. Kurt will combine 3 jars of paint into a new container to make a new color which he will name according to the following conditions:

Brun Y if the paint contains 2 jars of brown paint and no yellow
Brun X if the paint contains 3 jars of brown paint
Jaune X if the paint contains at least 2 jars of yellow
Jaune Y if the paint contains exactly 1 jar of yellow

What is the probability that the new color will be jaune?

A) 5/42
B) 37/42
C) 1/21
D) 4/9
E)5/9

I did.........

4/9*5/8*4/7 + 4/9*3/8*5/7 + 4/9*3/8*2/7
(Jaune Y) + (Jaune X) + (Jaune X)

= 41/126

??????

Answer is (b) 37/42
it will be jaune X if at least 2 yellow jars are there and jaune Y for a single yellow jar

so

p= 4c2.5c1/9c3 + 4c3.5c0/9c3 + 4c1.5c2/9c3

p= (30+4+40)/84 = 37/42


regards
badboys

[sumit82]

Originally Posted by badboys

Answer is (b) 37/42
it will be jaune X if at least 2 yellow jars are there and jaune Y for a single yellow jar

so

p= 4c2.5c1/9c3 + 4c3.5c0/9c3 + 4c1.5c2/9c3

p= (30+4+40)/84 = 37/42


regards
badboys

Plz let me know what is wrong with my approach.....

I have done the same way.... plz look into my solution and advice......

[Naresh Reddy]

Originally Posted by sumit82

Can someone help me with this, I can;t get where exactly I am going wrong.....


Q. Kurt, a painter, has 9 jars of paint 4 of which are yellow, 2 are red and the remaining jars are brown. Kurt will combine 3 jars of paint into a new container to make a new color which he will name according to the following conditions:

Brun Y if the paint contains 2 jars of brown paint and no yellow
Brun X if the paint contains 3 jars of brown paint
Jaune X if the paint contains at least 2 jars of yellow
Jaune Y if the paint contains exactly 1 jar of yellow

What is the probability that the new color will be jaune?

A) 5/42
B) 37/42
C) 1/21
D) 4/9
E)5/9

I did.........

4/9*5/8*4/7 + 4/9*3/8*5/7 + 4/9*3/8*2/7
(Jaune Y) + (Jaune X) + (Jaune X)

= 41/126

??????

another method..

p(jaune)+p(not jaune)=1

P(not jaune)= 5c3/9c3= 10/84
so p(jaune)= 1-10/84=74/84=37/42ANS.

[sumit82]

Originally Posted by Naresh Reddy

another method..

p(jaune)+p(not jaune)=1

P(not jaune)= 5c3/9c3= 10/84
so p(jaune)= 1-10/84=74/84=37/42ANS.

Guys Guys Guys

Lets not discuss alternatives here....

I really want to know what am I missing out on

I want to know where my answer is going wrong

Regards
Sumit

[popeye]

Originally Posted by sumit82

Guys Guys Guys

Lets not discuss alternatives here....

I really want to know what am I missing out on

I want to know where my answer is going wrong

Regards
Sumit

sumit dear,

your approach is wrong.

Consider JanuneY which is possible by exactly 1 yellow jar which can be chosen out of 4 in 4C1 ways. The remaining two jars can be selected out of 5 jars in 5C2 ways. This is a combined event.

what you are doing is multiplying seperate probabilities which is wrong. You are taking probabilty of yellow jar 4/9 which is wrong because that is probabilty of selecting just 1 jar. You have to take the other 2 jars into consideration togather.

Hence the probability of the event that JauneY is formed is 4C1*5C2/9C3 since 3 jars can be selected out of total 9 in 9C3 ways.

Similarly go for JauneX & you get the solution given by badboys

[made_for_iims]

Its a CSP GS Question.

Q. Ten Identical particles are moving randomly inside a closed box.what is the probability that any given point of time all the ten particles will be lying in the same half of the box?

[Naresh Reddy]

Originally Posted by made_for_iims

Its a CSP GS Question.

Q. Ten Identical particles are moving randomly inside a closed box.what is the probability that any given point of time all the ten particles will be lying in the same half of the box?

let us divide the box into two halves P & Q.
probability of a particle being in half P be x and that in Q be y
then x=y=1/2

now for 10 particles the probability follows binomial expansion (x+y)^10
in which a particular term [10cr * x^r* y^(10-r) ] refers to probability of "r "particlesin half P and the rest "10-r" particles in half Q..

now we want all 10 particles to be in one half i.e either P or Q

so probability reqd is
10c0 x^0 y^10+10c10 x^10 y^0=2*(1/2)^10= 1/2^9 ANS

[made_for_iims]

Originally Posted by Naresh Reddy

let us divide the box into two halves P & Q.
probability of a particle being in half P be x and that in Q be y
then x=y=1/2

now for 10 particles the probability follows binomial expansion (x+y)^10
in which a particular term [10cr * x^r* y^(10-r) ] refers to probability of "r "particlesin half P and the rest "10-r" particles in half Q..

now we want all 10 particles to be in one half i.e either P or Q

so probability reqd is
10c0 x^0 y^10+10c10 x^10 y^0=2*(1/2)^10= 1/2^9 ANS

I got diff sets of answers considering that we can divide the box into two halves in many ways....but still it is not comin closer to any options....

options are:-

1) 1/2 2) 1/5 3) 1/9 4) 1/11

Question was exactly this.......n i think options should be correct as well.....(As Its civil services Exam.....)

[rajsher]

Originally Posted by made_for_iims

I got diff sets of answers considering that we can divide the box into two halves in many ways....but still it is not comin closer to any options....

options are:-

1) 1/2 2) 1/5 3) 1/9 4) 1/11

Question was exactly this.......n i think options should be correct as well.....(As Its civil services Exam.....)

If i wanna take a guess, i will go with (3)...
I think if u cn tell us the right answer,then may be we can back track to find the right way to solve this question.

[made_for_iims]

Originally Posted by rajsher

If i wanna take a guess, i will go with (3)...
I think if u cn tell us the right answer,then may be we can back track to find the right way to solve this question.

No i dont know the answer.....
oops sorry
by mistake i wrote options wrong....
its 1/2 1/5 2/9 2/11

these options are 100% confirmed.......

It is troublin me more coz it was asked in general studies paper(not in the optional subject).....
so it should not be hard.... the other questions in paper were damn easy.....

Who knows,It may also come in CAT2005