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There are 7 b***s in a jar which have been numbered 0, 2, 3, 4, 6, 7, 9. Two b***s are drawn at random and the order in which the b***s are drawn does not matter. What is the probability of drawing a ball numbered with 2 and a ball numbered with 6?

 Feb 11, 2017
 #1
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The possible permuted sets of choosing 2 b***s from 7  = P(7,2)  = 42

 

But the only ones we are interested in are  {2,6} or {6,2}

 

So.....

 

2 / 42  =  1 / 21   ≈ 4.76%

 

 

cool cool cool

 Feb 11, 2017
 #2
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Thank you CPhill! 

Jeromex  Feb 11, 2017
 #3
avatar+129840 
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OK......no prob......

 

 

 

cool cool cool

 Feb 11, 2017

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