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# The following table shows the contents of a bag containing a variety

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Re: The following table shows the contents of a bag containing a variety [#permalink]
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Total objects = 17
Red objects = 7
Non_red Objects = 10

Probability of choosing at least one red = 1 - Probability of no red
= 1- $$\frac{10C3}{17C3}$$ = $$\frac{14}{17}$$

Option (D)

Bunuel wrote:
The following table shows the contents of a bag containing a variety of objects.

If three objects are drawn at random from the bag, then what is the probability that at least one of them is red?

A. 3/17
B. 45/136
C. 7/17
D. 14/17
E. 149/170

Project PS Butler

Attachment:
2020-09-07_17-53-17.png
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Re: The following table shows the contents of a bag containing a variety [#permalink]
1 - (10C3)/(17C3) = 1 - 3/17 = 14/17 (D)
What is challenging is manipulating these fractions without errors
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Re: The following table shows the contents of a bag containing a variety [#permalink]
1
Kudos
Bunuel wrote:
The following table shows the contents of a bag containing a variety of objects.

If three objects are drawn at random from the bag, then what is the probability that at least one of them is red?

A. 3/17
B. 45/136
C. 7/17
D. 14/17
E. 149/170

Project PS Butler

Attachment:
2020-09-07_17-53-17.png

Took some time to solve this - ~3 mins. My approach below -

Total blue objects = 8
Total red objects = 7
Total yellow objects = 2

No. of ways in which we can select any 3 objects = 17C3
No. of ways in which we can select 3 objects none of which is red = 10C3

Therefore probability of at least one red = (17C3 - 10C3) / (17C3) = 14/17

So D
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Re: The following table shows the contents of a bag containing a variety [#permalink]
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Probability when NONE is red = 10/17 * 9/16 * 8/15 = 3/17

Probability atleast one is RED = 1- 3/17 = 14/17.

So, I think D.
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Re: The following table shows the contents of a bag containing a variety [#permalink]
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Re: The following table shows the contents of a bag containing a variety [#permalink]
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