probability!!!!!!!!!!

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probability!!!!!!!!!!

by siddarthd2919 » Sat Mar 08, 2008 9:16 am
hi can some one help me with this one

a bag contains 3 red and 7 blue marbles.......2 marbles are drawn at random wat is the probability that ATLEAST 1 is blue?

a. 21/50
b. 3/13
c. 1/6
d. 1/12
e. 1/3

prob:2

a certain deck of cards has 2 blue,2 red,2 green and 2 yellow cards. if 2 cards are drawn at random wat is the probability that both are not blue?

a. 15/28
b. 1/4
c. 9/16
d. 1/32
e. 1/16



i dont have the qa........can some one post a brief explanation of these problems.......
:D

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Re: probability!!!!!!!!!!

by Stuart@KaplanGMAT » Sat Mar 08, 2008 3:21 pm
siddarthd2919 wrote:hi can some one help me with this one

a bag contains 3 red and 7 blue marbles.......2 marbles are drawn at random wat is the probability that ATLEAST 1 is blue?

a. 21/50
b. 3/13
c. 1/6
d. 1/12
e. 1/3
There are two ways we can use math to solve this: the fast way and the slow way!

The slow way is to recognize that there are 3 scenarios that match the "at least one blue" requirement:

R B
B R
B B

So, one way we could solve the problem is to calculate the probability of each scenario occoring and then, since we want scenario 1 OR scenario 2 OR scenario 3, to add those probabilities together.

Accordingly:

R B = (3/10)(7/9) = 21/90
B R = (7/10)(3/9) = 21/90
B B = (7/10)(6/9) = 42/90

So, the probability of getting AT LEAST 1 blue is (21+21+42)/90 = 84/90 = 14/15

(So either you've stated the question wrong or the choices are messed up. In fact, just by estimating we should be able to conclude that if the probabiliy that 1 marble is blue is > 50%, the probability that at least one out of two being blue should be even higher, so none of those choices could possibly be correct.)

The quicker method to solve this type of question is to know that the sum of all possible events is 1 and to recognize that if we want AT LEAST 1 blue, the only event we don't want is:

R R

So, we instead of adding up the 3 events we do want to happen, we can instead use the formula:

Prob (things we want) = 1 - Prob (things we don't want)

or

Prob (at least 1 blue) = 1 - Prob (RR)
Prob (at least 1 blue) = 1 - (3/10)(2/9) = 1-6/90 = 84/90 = 14/15


a certain deck of cards has 2 blue,2 red,2 green and 2 yellow cards. if 2 cards are drawn at random wat is the probability that both are not blue?

a. 15/28
b. 1/4
c. 9/16
d. 1/32
e. 1/16
The first problem with this question is ambiguity. "What is the probability that both are not blue" could be interpreted as "neither is blue" or as "at least one isn't blue". Technically, by placing "both" where it is in the sentence, the question is asking for the probability that NEITHER is blue. The latter interpretation would be more accurate if the question had asked "What is the probability that not both are blue?"

Since the "neither" interpretation is the correct one, let's solve it that way and see what happens.

As we saw above, we can attack this two ways - we can either add up all the things we do want (which would take insanely long in this question) or use the 1 - (what we don't want) method.

The only thing we don't want is BB, so:

Probability (not BB) = 1 - Prob (BB) = 1 - (2/8)(1/7) = 1 - 2/56 = 54/56 = 27/28

Which again isn't one of the choices. The fact that we don't have a match leads me to a few subsidiary conclusions:

(1) the writer may have intended the second possible interpretation;
(2) the writer may have intended the draw to be "with replacement", which would change the results (by not specifying with or without replacement, we should assume that the draws are taking place simultaneously, i.e. you pull out 2 cards at once, so there's no replacement); and/or
(3) the answer choices provided are wrong

and, coupled with the first question (which I assume comes from the same place), one big conclusion:

(1) the source of these questions is NOT to be trusted.

What is the source? It's always a good idea to post that information, so we can better evaluate the validity of the questions. These two are definitely GMAT-style questions, they just seem to be horribly miswritten (or possibly misquoted).
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by siddarthd2919 » Sat Mar 08, 2008 5:12 pm
THANKS STUART ........ I FOUND THIS QUESTION IN THE YAHOO GROUPS....... I TOO HAD THE FEELING THAT THE QUESTIONS WHERE WRONG......BUT DIDNT HAVE THE CONFIDENCE, TO LEAVE THEM AS WRONG. :D

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The probability of not drawing a blue card on the first draw is

= (number of cards that are not blue) / (total number of cards)
= 6/8

The probability of not drawing a blue card on the second draw is
= (number of cards that are not blue) / (total number of cards)
= 5/7

= (6/8) * (5/7)
= 30/56
= 15/28

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Sorry some smileys got clicked...

by sssmoorthy » Sat Aug 16, 2008 6:54 am
The probability of not drawing a blue card on the first draw is

= (number of cards that are not blue) / (total number of cards)
= 6/8

The probability of not drawing a blue card on the second draw is
= (number of cards that are not blue) / (total number of cards)
= 5/7

= (6/8) * (5/7)
= 30/56
= 15/28 :?

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by ronsom » Tue Apr 27, 2010 12:11 am
I think getting 27/28 for neither of the cards being blue is a little ambiguous!
It is not possible for such an event to occur with a probability of 1 (27/28 is almost equal to 1)
.. so i think the 15/28 answer is more convincing..

however... now i'm confused as to which approach should be followed when similar questions are encountered!
Staurt followed an approach wherein he did:-
P(event does not occur) = 1 - P(event occurs)

others who concluded to 15/28 as the answer followed an approach wherein they calculated the Probability of an event not occurring in a different way..

which approach to follow when.. is my question

Thanks