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Bunuel
A consignment of eight washing machines contains two fully-automatic washing machines and six semi-automatic washing machines. If two washing machines are to be chosen at random from this consignment, what is the probability that at least one of the two washing machines chosen will be a fully-automatic?


(A) 1/5

(B) 1/4

(C) 3/14

(D) 13/28

(E) 17/28

P of semi auto ; 6/8 * 5/7 = 15/28
P of auto = 1-15/28 ; 13/28
IMO D
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I thought the same approach too but I am not clear about considering order of picking the washing machines. Why do I need to multiply by 2 in second case.
Could you explain it or send links to clear this doubt?

By different approach I am getting correct answer.
(2c1x6c1 + 2c2)/8c2= 13/28


chetan2u
Bunuel
A consignment of eight washing machines contains two fully-automatic washing machines and six semi-automatic washing machines. If two washing machines are to be chosen at random from this consignment, what is the probability that at least one of the two washing machines chosen will be a fully-automatic?


(A) 1/5

(B) 1/4

(C) 3/14

(D) 13/28

(E) 17/28


Another way is through probability..
AT LEAST 1 means 2 cases
(I) The first and second both are fully-auto = \(\frac{2}{8}*\frac{1}{7}=\frac{1}{28}\)
(II) Only one is fully auto = \(\frac{2}{8}*\frac{6}{7}=\frac{6}{28}\), and as we can have that one auto picked up in first pick or second pick, multiply by 2 = \(2*\frac{6}{28}\)

Total = \(\frac{1}{28}+2*\frac{6}{28}=\frac{13}{28}\)

D
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Bunuel
A consignment of eight washing machines contains two fully-automatic washing machines and six semi-automatic washing machines. If two washing machines are to be chosen at random from this consignment, what is the probability that at least one of the two washing machines chosen will be a fully-automatic?


(A) 1/5

(B) 1/4

(C) 3/14

(D) 13/28

(E) 17/28

We can use the formula:

1 - P(no fully-automatic washing machines selected) = P(at least one fully-automatic washing machine selected)

P(no fully-automatic washing machines selected) = 6/8 x 5/7 = 30/56 = 15/28.

P(at least one fully-automatic washing machine selected = 1 - 15/28 = 13/28.

Alternate Solution:

2 washing machines can be chosen out of 8 washing machines in 8C2 = 8!/(2!*6!) = (8*7)/2 = 28 ways.

One fully automatic and one semi-automatic washing machine can be chosen is 2C1 * 6C1 = 2 * 6 = 12 ways.

Two fully automatic washing machines can be chosen in 2C2 = 1 way.

Thus, the requirement of “selection includes at least one fully automatic washing machine” can be satisfied in 12 + 1 = 13 ways.

So, the probability of such a selection is 13/28.

Answer: D
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Thanks for the great explanation, chetan2u
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gvij2017
I thought the same approach too but I am not clear about considering order of picking the washing machines. Why do I need to multiply by 2 in second case.
Could you explain it or send links to clear this doubt?

By different approach I am getting correct answer.
(2c1x6c1 + 2c2)/8c2= 13/28


chetan2u
Bunuel
A consignment of eight washing machines contains two fully-automatic washing machines and six semi-automatic washing machines. If two washing machines are to be chosen at random from this consignment, what is the probability that at least one of the two washing machines chosen will be a fully-automatic?


(A) 1/5

(B) 1/4

(C) 3/14

(D) 13/28

(E) 17/28


Another way is through probability..
AT LEAST 1 means 2 cases
(I) The first and second both are fully-auto = \(\frac{2}{8}*\frac{1}{7}=\frac{1}{28}\)
(II) Only one is fully auto = \(\frac{2}{8}*\frac{6}{7}=\frac{6}{28}\), and as we can have that one auto picked up in first pick or second pick, multiply by 2 = \(2*\frac{6}{28}\)

Total = \(\frac{1}{28}+2*\frac{6}{28}=\frac{13}{28}\)

D


Hi,
It is multiplied by 2, as there can be two ways to pick one fully auto and one semi auto, that is 2 orders--- FS or SF
That is why we multiply by 2
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Bunuel
A consignment of eight washing machines contains two fully-automatic washing machines and six semi-automatic washing machines. If two washing machines are to be chosen at random from this consignment, what is the probability that at least one of the two washing machines chosen will be a fully-automatic?


(A) 1/5

(B) 1/4

(C) 3/14

(D) 13/28

(E) 17/28

This question is a pretty straightforward "1 minus" probability question and the approach has been covered in the posts above, so I won't duplicate. I just wanted to take a minute to give a little tip about how to avoid silly mistakes that sometimes trap students.

15/28 doesn't happen to be an answer choice here, but I'm guess that it would be an option if this same question were written by GMAC. In that case, I'm sure you can imagine doing the correct math most of the way through the question, simply forgetting to subtract from 1, and choosing an incorrect answer choice. Not because you didn't know how to do it, but because there's some step you knew you were supposed to do and then just forgot. This question is a super easy example, but you've probably seen some that have many more steps and then end with one final twist. Maybe a "1 minus," maybe the question asks for the area of a square and you spend a bunch of time calculating the side but forget to square it to find the area, maybe there's some complex math that allows you to calculate the profit per month but the question asks for the profit per quarter and you forget to multiply by three. Anyway, a tonnnnn of ways for them to set this stuff up. Here's my tip: As you're reading the question stem, thinking about how to set up the math on your whiteboard, and notice that there's a last-step adjustment that's going to be necessary once you get allllllmost to the finish line, flip your computer mouse over. When you finish your calculations and go to click an answer, now you have something to remind you to do that last step if you haven't done so already. Getting the best score you can is sometimes as much about protecting yourself from dumb mistakes as it is about knowing how to do the math!
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The simplest way to do this question would be considering none of the machines are auto and minus that from 1 . that plan would give as probabilities of 6/8*5/7=30/56 or 15/28

Finally 1-15/28= (28-15)/28 = 13/28

D)

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