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

Question

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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Official Answer

D

rajatchopra1994 · Answer

Explanation:

total objects = 17
Red objects = 7
Non Red Objects = 17 - 7 = 10

Probability of no red in 3 draws = 10/17 * 9/16 * 8/15
= 3/17

Probability of at least one red = 1 - 3/17 = 14/17

IMO-D

asishron29181 · Answer

Total items: 17
Non red items = 10
Probability of selecting atleast one red item = 1 - Probability of selecting none red items
Total ways of selecting = 17 C 3 = 17*16*15 / 3*2*1 = 680
Total ways of selecting none red items = 10C3 = 10*9*8 / 3*2 = 120

Required probability = (1 - 120/ 680 )
= 14/17

IMO D

sumitkrocks · Answer

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)

kantapong · Answer

1 - (10C3)/(17C3) = 1 - 3/17 = 14/17 (D) What is challenging is manipulating these fractions without errors

thelonghalloween · Answer

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

TarunKumar1234 · Answer

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

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