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probability

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Junior | Next Rank: 30 Posts
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by ttv » Mon Jan 07, 2008 12:27 pm
A bowl contains one marble labeled 0, one marble labeled 1, one marble labeled 2, one marble labeled 3. the bowl contains no other objects. If two marbles are drawn randomly without replacement, what is the probility that they willa dd up to 3 ?

(A)1/12
(B)1/8
(C)1/6
(D)1/4
(E)1/3

-thanx
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by cris » Mon Jan 07, 2008 1:10 pm
E??

Total possibilities: 6

Possibilities that sum up 3: 2

2/6=1/3
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by ttv » Mon Jan 07, 2008 1:59 pm
how u get this answer. . . m so sorry but m not good in probability . . . i find it lil difficult. so please help me out ?


thanx
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by ttv » Mon Jan 07, 2008 2:03 pm
A housing subdividion contains only 2 types of homes : ranch-style homes and townhomes. There are twice as many townhomes as ranch-homes. There are 3 times as many townhomes with pools than without pools. What is the probability that a home selected at random from the subdivision will be a townhome with a pool ?

(A)1/6
(B)1/5
(C)1/4
(D)1/3
(E)1/2


-thanx
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by cris » Mon Jan 07, 2008 2:56 pm
C?? 1/4

TTV please tell me if these are the correct answers because otherwise I really should not be trying to explain anything to you...because it would be wrong
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by camitava » Mon Jan 07, 2008 10:34 pm
ttv,
In response to ur first Qs, I would like to take the following approach -
We can get 3 if we chose the following marbles -
(3,0) and (1,2) -> So number of chances = 2
But w/o any condition, we can take 2 marbles out of 4 in 4C2 = 6 ways [Here C -> Combination]
So probability = 2/6 = 1/3
Now according to ur second Qs,
Let, ranch-style homes = R and townhomes = T
and townhomes with pools = WT
and townhomes without pools = WoT

According to the problem, WT = 3WoT and T = 2R
So we can say WP + 3WoT = T
And WT + WoT = 2R
or 4WoT = 2R
or R = 2WoT
Total no of houses = T + R = 4WoT + 2WoT = 6WoT
And we have to select a WT from 6WoT.
So the probability = 3WoT/6WoT = 1/2 - So IMO E.

But what are the OAs? Please specify ...
Correct me If I am wrong


Regards,

Amitava
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Re: probability

by ChesireCat » Tue Jan 08, 2008 5:35 am
ok...look at it this way... you have to pick 2 balls.

now when u pick the first ball.... it may be either 0,1,2 or 3.

If 0 then we need next ball numbered 3 out of 3 balls to sum up as 3...
probability of getting ball with #3 is 1\3

Similarly
If 1 then we need next ball 2 to sum up as 3...
probability of getting ball with #2 is 1\3


Similarly
If 2 then we need next ball 1 to sum up as 3...
probability of getting ball with #1 is 1\3


Similarly
If 3 then we need next ball 0 to sum up as 3...
probability of getting ball with #0 is 1\3


therefore the probablity in either case is 1\3.
ttv wrote: A bowl contains one marble labeled 0, one marble labeled 1, one marble labeled 2, one marble labeled 3. the bowl contains no other objects. If two marbles are drawn randomly without replacement, what is the probility that they willa dd up to 3 ?

(A)1/12
(B)1/8
(C)1/6
(D)1/4
(E)1/3

-thanx
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by Bronson » Tue Jan 08, 2008 8:05 am
ttv wrote:A housing subdividion contains only 2 types of homes : ranch-style homes and townhomes. There are twice as many townhomes as ranch-homes. There are 3 times as many townhomes with pools than without pools. What is the probability that a home selected at random from the subdivision will be a townhome with a pool ?

Maybe someone will find this method helpful:

2R=T
Probability to pick R is 1/3 and for T is 2/3. They can be T=40 and R=20.

Tp=3Twp Probability to pick Tp is 3/4 and for Twp is 1/4.

Probability to pick T and Tp is 2/3 * 3/4 = 1/2, or if T=40, than Tp is 30 and Twp is 10, and than 30 out of 60 is 1/2
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by preciousrain7 » Tue Jan 08, 2008 3:29 pm
camitava wrote:ttv,
In response to ur first Qs, I would like to take the following approach -
We can get 3 if we chose the following marbles -
(3,0) and (1,2) -> So number of chances = 2
But w/o any condition, we can take 2 marbles out of 4 in 4C2 = 6 ways [Here C -> Combination]
So probability = 2/6 = 1/3
...
Hi Camitava, can you explain to me how you get 6 ways when doing 4C2? THANKS!
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by camitava » Tue Jan 08, 2008 3:33 pm
preciousrain7, 4C2 means we taking a combination of 2 out of 4. The formula is - mCn = m!/((m - n)! * n!). Now put m = 4 and n = 2 and u will get 6.
Correct me If I am wrong


Regards,

Amitava
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by preciousrain7 » Tue Jan 08, 2008 3:46 pm
camitava wrote:preciousrain7, 4C2 means we taking a combination of 2 out of 4. The formula is - mCn = m!/((m - n)! * n!). Now put m = 4 and n = 2 and u will get 6.
THANKS!!!!!!!!
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by moreonalps » Tue Jan 08, 2008 6:59 pm
On the first one, the question says "what is the probility that they will add up to 3"

Don't you think number of favorable cases are [0,1],[0,2][0,3][1,2] which means probability should be 4/6=>2/3..something appears to be incorrect..question/answer choices or my interpretation..
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by camitava » Wed Jan 09, 2008 8:56 pm
moreonalps, obviously I understood sum up to 3 means only sum of the two marbles will be 3 only! If not, then what u have mentioned is the best way to solve the problem ... TTV, request you to mention the OA. And if u can focus some light on it, please do so ...
Correct me If I am wrong


Regards,

Amitava
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Housing problem

by roy_priya » Fri Jan 18, 2008 2:59 pm
An easy way to solve this problem is plugging in numbers.

Let's consider number of townhouses without pool = 1

With pool = 3 x 1 = 3

Total town houses = 3 + 1 = 4

Ranch houses ( as per the the problem) = 4/2 = 2

Total number of houses = 4+2 = 6

Probability of picking a pool house = 3/6 = 1/2
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by simplyjat » Fri Jan 18, 2008 8:36 pm
Can we just stick to one question per thread....
Threads are not in short supply and not going to be extinct...
simplyjat
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