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# On a store counter are exactly two boxes containing only purple marble

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On a store counter are exactly two boxes containing only purple marble [#permalink]

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18 May 2015, 19:59
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Question Stats:

71% (01:16) correct 29% (01:05) wrong based on 325 sessions

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On a store counter are exactly two boxes containing only purple marbles and yellow marbles. Box A has 30 purple marbles and 20 yellow marbles; box B has 15 purple marbles and 35 yellow marbles. If Melanie randomly selects one of the boxes and then randomly selects one marble from the box, what is the probability that the marble selected will be purple?

(A) 3/20
(B) 3/10
(C) 9/20
(D) 3/5
(E) 9/10
[Reveal] Spoiler: OA

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Re: On a store counter are exactly two boxes containing only purple marble [#permalink]

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18 May 2015, 20:11
We have 30+15 =45 purple marbles
Box A and box B total marbles = 100 marbles

Probability that the marble selected will be purple =$$\frac{45}{100}$$= $$\frac{9}{20}$$

Hence, answer is C, but is this the right way to solve this problem?

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Re: On a store counter are exactly two boxes containing only purple marble [#permalink]

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18 May 2015, 22:02
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rlitagmatstudy wrote:
We have 30+15 =45 purple marbles
Box A and box B total marbles = 100 marbles

Probability that the marble selected will be purple =$$\frac{45}{100}$$= $$\frac{9}{20}$$

Hence, answer is C, but is this the right way to solve this problem?

Hi rlitagmatstudy,

I would suggest you approach it this way:

The question asks us to select a purple marble from either of the boxes. Here we need to select a box first(either A or B) and then select the marble from the box. So these events are AND events i.e. both need to be done to select a purple marble from the box. For an AND event we multiply the probability of the events. Hence the probability equation can be written as:

P(Selecting a purple marble) = P(selecting a box) * P(selecting purple marble from the selected box)

The probability of selecting either of the box is the same = $$\frac{1}{2}$$

$$= \frac{1}{2} * \frac{30}{50} + \frac{1}{2} * \frac{15}{50} = \frac{9}{20}.$$

Doing it this way will help you take care of the difference in probability of selecting the box as well as difference in probability of selecting the marble. You go the right answer by dividing the total purple marbles by total number of marbles as there were same number of marbles in both the boxes and the probability of selecting either of the box was same.

Another example

Box A - 20 Purple and 20 yellow marbles
Box B - 15 purple and 35 yellow marbles

Using your method will give us the P(selecting a purple marble) = $$\frac{35}{90} = \frac{7}{18}$$.

However the right answer would be P(selecting a purple marble) = $$\frac{1}{2} * \frac{20}{40} + \frac{1}{2} * \frac{15}{50}= \frac{2}{5}.$$

Another variation of this question can be where the probability of selecting the boxes may not be the same. Hence you can't take the ratio of total purple marbles to the total marbles to arrive at the probability of selecting a purple marble from both the boxes.

Hope this helps

Regards
Harsh
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On a store counter are exactly two boxes containing only purple marble [#permalink]

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02 Nov 2015, 20:39
I think this problem is much simpler than it seems: as long as Melanie is doing one action at a time, i.e. picks randomly one box and drags randomly one ball and is not doing 2 actions or considering which of the two boxes to prefer - these 2 boxes from the probability standpoint are one large box!

The above means that we simply sum all the balls and see the individual probability of picking one purple which is:

$$\frac{45}{100}$$ = $$\frac{9}{20}$$

is my reasoning correct? what if there were 3 boxes?
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On a store counter are exactly two boxes containing only purple marble [#permalink]

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19 Jun 2016, 02:46
A rather basic question here,

Aren't we suppose to subtract the overlap when calculating the P(A or B)?

Here we only do P(A) + P(B) and that's it. Is it because the P(selecting any box) is the same? So it's basically one big box with all the marbles in it?

Or is it because the marble can't be of both colors at the same time?..

It seems that these basic concepts are so easy until you find them in a real example!..

Thanks!

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Re: On a store counter are exactly two boxes containing only purple marble [#permalink]

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30 Jul 2016, 15:24
iliavko wrote:
A rather basic question here,

Aren't we suppose to subtract the overlap when calculating the P(A or B)?

Here we only do P(A) + P(B) and that's it. Is it because the P(selecting any box) is the same? So it's basically one big box with all the marbles in it?

Or is it because the marble can't be of both colors at the same time?..

It seems that these basic concepts are so easy until you find them in a real example!..

Thanks!

The general formula for OR probability is P(A or B)=P(A) + P(B) - P(A and B). However P(A and B), which is the overlap that you mentioned, is equal to zero in this question. In other words, we assign a value to P(A and B) ONLY when P(A) and P(B) share an outcome.

For example: A die is rolled. What is the probability that the number is even or less than 4?

Probability that the number is even: this includes {2,4,6} therefore, P(A) = 3/6
Probability that the number is less than four: this includes {1,2,3} therefore, P(B) = 3/6

If you notice both P(A) and P(B) are sharing the number 2 in their outcome. therefore, P(A and B) = 1/6

P(A or B)= (3/6) + (3/6) - (1/6) = 5/6

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Re: On a store counter are exactly two boxes containing only purple marble [#permalink]

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13 Sep 2016, 13:05
Hi,

I did it this way :
purple marble were chosen either from A or B. P (selecting purple from A ) : 30/50 ,similarly from B : 15/50,
p(A) + P(B) = 9/10. Can someone please explain why this is incorrect.

Regards
Megha

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Re: On a store counter are exactly two boxes containing only purple marble [#permalink]

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15 Sep 2016, 06:34
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rlitagmatstudy wrote:
On a store counter are exactly two boxes containing only purple marbles and yellow marbles. Box A has 30 purple marbles and 20 yellow marbles; box B has 15 purple marbles and 35 yellow marbles. If Melanie randomly selects one of the boxes and then randomly selects one marble from the box, what is the probability that the marble selected will be purple?

(A) 3/20
(B) 3/10
(C) 9/20
(D) 3/5
(E) 9/10

We are given that box A has 30 purple marbles and 20 yellow marbles and that box B has 15 purple marbles and 35 yellow marbles. We need to determine the probability, when selecting one marble from the box, of selecting a purple marble. Since we have two boxes, A and B, there are multiple scenarios to account for when selecting the marbles. We must account for the probability of first selecting either box and then secondly for selecting a purple marble. Let’s start with Box A.

P(selecting box A) = ½

P(selecting a purple marble in box A) = 30/50 = ⅗

Thus, the probability of selecting a purple marble from box A is ½ x ⅗ = 3/10

Next we can determine the probability of selecting a purple marble from box B.

P(selecting box B) = ½

P(selecting a purple marble in box B) = 15/50 = 3/10

Thus, the probability of selecting a purple marble from box B is ½ x 3/10 = 3/20

Now we can determine the probability of selecting a purple marble from box A or box B:

3/10 + 3/20 = 6/20 + 3/20 = 9/20.

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Re: On a store counter are exactly two boxes containing only purple marble [#permalink]

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29 Oct 2017, 15:00
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Re: On a store counter are exactly two boxes containing only purple marble   [#permalink] 29 Oct 2017, 15:00
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