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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




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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


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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]
2
Kudos
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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