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DS - Probability

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DS - Probability

by karthikpandian19 » Wed Jul 04, 2012 9:16 pm
Each of the twelve articles of clothing in a laundry basket is either a sweater or a t-shirt, and each is either blue, black, or yellow. If one article of clothing is randomly selected from the laundry basket, what is the probability that the article selected will either be a sweater or blue?

The laundry basket contains no blue sweaters.
The probability that the article of clothing will be a sweater minus the probability that it will be blue is 0.5
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by eagleeye » Thu Jul 05, 2012 12:06 am
karthikpandian19 wrote:Each of the twelve articles of clothing in a laundry basket is either a sweater or a t-shirt, and each is either blue, black, or yellow. If one article of clothing is randomly selected from the laundry basket, what is the probability that the article selected will either be a sweater or blue?
We are told that the basket has 12 articles and that
a. Each article is either a sweater or a t-shirt.
b. Each article is either blue, or black, or yellow.

We need to find the probability that a randomly selected articles is either a sweater or blue. Since we know the overall number is 12, if we can find how many articles of clothing are blue or are sweaters or both, we will be good. I am going to take this route for ease of explanation.

With that in mind, let's look at the statements:

The laundry basket contains no blue sweaters.
This tells us that no blue articles are sweaters. In other words, all blue clothes are t-shirts.
This doesn't tell us anything about how many sweaters and blues there are. Insufficient.

The probability that the article of clothing will be a sweater minus the probability that it will be blue is 0.5
This tells us that the number of non-blue sweaters is greater than that of blue t-shirts by (0.5*12) = 6.

(Sweater = Blue Sweater + Not Blue Sweater ; Blue = Blue Sweater + Blue T-shirt
so, Sweater - Blue = Not Blue Sweater + Blue t-shirt)

It doesn't talk about how many are sweaters, or how many are blues. Insufficient.

1 and 2 combined tell us that there are no blue sweaters. So, the second statement modifies to:
The number of sweaters is greater than that of blues by 6.

Now we can have (0 blues, 6 sweaters) OR (1 blue, 7) sweaters, OR (2 blues, 8) sweaters, all the way up to (6 blues, 6 sweaters). So, the probability of blues OR sweaters lies anywhere from (0+6)/12 to (6+6)/12 => between 0.5 and 1. So together, we still don't have a fixed number for probability. Insufficient.

Hence the final answer is E.

Let me know if this helps :)
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