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Probability

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Probability

by HarvardDreamin » Fri Mar 21, 2008 9:00 am
Q2) A shipment of 8 television sets contains 2 black-and-white sets and 6color sets. If 2 television sets are to be chosen at random from this shipment, what is the probability that at least 1 of the 2 sets chosen will be a black-and-white set?

A. 1/7 B. ¼ C. 5/14 D. 11/28 E. 13/26
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by xilef » Fri Mar 21, 2008 10:10 am
probability that at least 1 of the 2 sets chosen will be a black-and-white set means that either one or both can be black-and-white.

First one yes, second no - 1/4 * 6/7 = 6/28
First one no, second yes - 3/4 * 2/7 = 6/28
Both are - 1/4 * 1/7 = 1/28

Now add all three scenarios:

6/28+6/28+1/28=13/28

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by musicdaemon » Fri Mar 21, 2008 10:45 am
i also think the ans is copied wrong,

i used the combinatorics to solve,

at least 1 black & white set is to be selected, i.e either one or both the sets are B&W.

Suppose, case is to select none of the B&W sets,

n(E)= number of events of two color sets being selected
= 6C2 = 15
n(S)= number of events of selecting any two sets
= 8C2 = 28

probability of selecting exclusively two color sets,

n(P) = (n(E))/(n(S))= 15/28

let n(U)=any two sets are selected=1

Therefore, for at least one B&W set to be selected

n(P') = 1-n(P)
= 1-(15/28) = 13/28............................... Ans.(xilef already solved)
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by HarvardDreamin » Fri Mar 21, 2008 11:41 am
Had the same query when I first solved it and thought I was incorrect.
The answers are exactly the same as in the question. I got it from the Sets package of questions. Im guessing either we're all missing something or the question is fundamentally flawed.
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by beeparoo » Sun May 11, 2008 5:36 pm
xilef wrote:probability that at least 1 of the 2 sets chosen will be a black-and-white set means that either one or both can be black-and-white.

First one yes, second no - 1/4 * 6/7 = 6/28
First one no, second yes - 3/4 * 2/7 = 6/28
Both are - 1/4 * 1/7 = 1/28

Now add all three scenarios:

6/28+6/28+1/28=13/28

Were the answers copied correctly?
What is the Answer according to the Set; did the Set come with an answer at all?

To me, the question stem does not distinguish between the order of selecting a B&W set first, or, second -- only that it would be selected AT LEAST once. While I agree with your approach in that both B&W sets may be selected too, I am not fully convinced that you need to account for the following:

"First one yes, second no"
"First one no, second yes"

Wouldn't the probability of one B&W and one colour (in no particular order) suffice?

Using this approach leads to answer B.

What do you think?
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