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few HOT and few NOT

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by sanju09 » Wed Feb 03, 2010 11:50 pm
There are cookies in a plate, few HOT and few NOT. A child each time randomly picks a single cookie out from the plate and eats it if it's HOT, or leaves the plate forever if it's NOT. What is the probability that the child will be able to eat at least 3 cookies before leaving the plate forever?

(1) Probability that child won't be able to eat any cookie before leaving the plate forever is 8/15.

(2) Probability that the child will be able to eat at least 2 cookies before leaving the plate forever is 11/15.








[spoiler]The label "few HOT and few NOT" clicked in mind this morning only, while doing a bathroom concert. Ardently, decided to make a question titled that. How is it? Don't know. In fact, haven't even given a single shot to it till now. Be first to tell what's wrong in it.[/spoiler]
The mind is everything. What you think you become. -Lord Buddha



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by harsh.champ » Thu Feb 04, 2010 2:23 am
sanju09 wrote:There are cookies in a plate, few HOT and few NOT. A child each time randomly picks a single cookie out from the plate and eats it if it's HOT, or leaves the plate forever if it's NOT. What is the probability that the child will be able to eat at least 3 cookies before leaving the plate forever?

(1) Probability that child won't be able to eat any cookie before leaving the plate forever is 8/15.

(2) Probability that the child will be able to eat at least 2 cookies before leaving the plate forever is 11/15.








[spoiler]The label "few HOT and few NOT" clicked in mind this morning only, while doing a bathroom concert. Ardently, decided to make a question titled that. How is it? Don't know. In fact, haven't even given a single shot to it till now. Be first to tell what's wrong in it.[/spoiler]
____________________
Statement 1:-It means that the first cookie that the child picks is a NOT cookie.
Hence the ratio of NOT:HOT = 8:7 (since 8/15 is the probability)
Now,P(at least 3 cookies)=1-P(0 cookie+1 cookie+2cookies)
P(0 cookie)=8/15 [1 NOT]
P(1 cookie)=(7/15)*(8/14) [1 HOT,1 NOT]
P(2 cookies)=(7/15)*(6/14)*(8/13) [1 HOT,1 HOT ,1 NOT]
Hence,the answer can be arrived at using the upper equation.

Statement 2:1-[P(0) + P(1)]=P[at least 2 cookies]=11/15
Now,P(at least 3 cookies)=1-P(0 cookie+1 cookie+2cookies)=1-P(0 cookie+1 cookie)-P(2cookies)
=11/15-P(2 cookies)
When we consider the no. of NOT COOKIES=x and HOT COOKIES=y,we arrive at a biquadratic equation .
[(x^2+2xy-x)]/[(x+y)(x+y-1)]=4/15
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by ajith » Thu Feb 04, 2010 2:39 am
sanju09 wrote:There are cookies in a plate, few HOT and few NOT. A child each time randomly picks a single cookie out from the plate and eats it if it's HOT, or leaves the plate forever if it's NOT. What is the probability that the child will be able to eat at least 3 cookies before leaving the plate forever?

(1) Probability that child won't be able to eat any cookie before leaving the plate forever is 8/15.

(2) Probability that the child will be able to eat at least 2 cookies before leaving the plate forever is 11/15.








[spoiler]The label "few HOT and few NOT" clicked in mind this morning only, while doing a bathroom concert. Ardently, decided to make a question titled that. How is it? Don't know. In fact, haven't even given a single shot to it till now. Be first to tell what's wrong in it.[/spoiler]
IMO, E
Since neither 1 or 2 will help us determine the number of HOT and NOT buns
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by sanju09 » Thu Feb 04, 2010 2:44 am
IMO, E
Since neither 1 or 2 will help us determine the number of HOT and NOT buns
When we set up two different equations in two variables, we sometime don't get to a unique set of answers. I'm not sure what has happened in this case.
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by harsh.champ » Thu Feb 04, 2010 5:27 am
ajith wrote:
sanju09 wrote:There are cookies in a plate, few HOT and few NOT. A child each time randomly picks a single cookie out from the plate and eats it if it's HOT, or leaves the plate forever if it's NOT. What is the probability that the child will be able to eat at least 3 cookies before leaving the plate forever?

(1) Probability that child won't be able to eat any cookie before leaving the plate forever is 8/15.

(2) Probability that the child will be able to eat at least 2 cookies before leaving the plate forever is 11/15.








[spoiler]The label "few HOT and few NOT" clicked in mind this morning only, while doing a bathroom concert. Ardently, decided to make a question titled that. How is it? Don't know. In fact, haven't even given a single shot to it till now. Be first to tell what's wrong in it.[/spoiler]
IMO, E
Since neither 1 or 2 will help us determine the number of HOT and NOT buns
____________________________-
I doubt that,ajith.
Just see the solution posted by me above.
Using only the statement 1,I have arrived at the solution.
Hence,E is definitely not the answer.

Taken from my solution:-""Now,P(at least 3 cookies)=1-P(0 cookie+1 cookie+2cookies)
P(0 cookie)=8/15 [1 NOT]
P(1 cookie)=(7/15)*(8/14) [1 HOT,1 NOT]
P(2 cookies)=(7/15)*(6/14)*(8/13) [1 HOT,1 HOT ,1 NOT]
Hence,the answer can be arrived at using the upper equation.""

Hence,P(at least 3 cookies)=1-[(8/15) + ((7/15)*(8/14)) + ((7/15)*(6/14)*(8/13))].
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by ajith » Thu Feb 04, 2010 5:31 am
harsh.champ wrote: Statement 1:-It means that the first cookie that the child picks is a NOT cookie.
Hence the ratio of NOT:HOT = 8:7 (since 8/15 is the probability)
Now,P(at least 3 cookies)=1-P(0 cookie+1 cookie+2cookies)
P(0 cookie)=8/15 [1 NOT]
P(1 cookie)=(7/15)*(8/14) [1 HOT,1 NOT]
P(2 cookies)=(7/15)*(6/14)*(8/13) [1 HOT,1 HOT ,1 NOT]
Hence,the answer can be arrived at using the upper equation.

Statement 2:1-[P(0) + P(1)]=P[at least 2 cookies]=11/15
Now,P(at least 3 cookies)=1-P(0 cookie+1 cookie+2cookies)=1-P(0 cookie+1 cookie)-P(2cookies)
=11/15-P(2 cookies)
When we consider the no. of NOT COOKIES=x and HOT COOKIES=y,we arrive at a biquadratic equation .
[(x^2+2xy-x)]/[(x+y)(x+y-1)]=4/15
You assume that there is only 15 cookies - what is this assumption based on?

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