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What can be the quickest approach to this question:-
A bag contains 3 Blue 5 Green 4 yellow balls 3 balls are drawn at random without replacement. What is the probability that exactly 2 of the 3 balls will be Blue.?
The approach i used is very lengthy and I think not correct or advisable
I used 1-(Prob of All Blue + Prob of no Blue + Prob of One Green)
Thanks
Posted from my mobile device
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What can be the quickest approach to this question:-
A bag contains 3 Blue 5 Green 4 yellow balls 3 balls are drawn at random without replacement. What is the probability that exactly 2 of the 3 balls will be Blue.?
The approach i used is very lengthy and I think not correct or advisable
I used 1-(Prob of All Blue + Prob of no Blue + Prob of One Green)
Thanks
Posted from my mobile device
Show more
The correct way would be Ways in which 2 are blue 3C2 and the third could be any of 5+4 or 9..3C2*9C1=3*9=27 Total ways = (3+5+4)C3=12C3=12*11*10/3*2=220
What can be the quickest approach to this question:-
A bag contains 3 Blue 5 Green 4 yellow balls 3 balls are drawn at random without replacement. What is the probability that exactly 2 of the 3 balls will be Blue.?
The approach i used is very lengthy and I think not correct or advisable
I used 1-(Prob of All Blue + Prob of no Blue + Prob of One Green)
Thanks
Posted from my mobile device
The correct way would be Ways in which 2 are blue 3C2 and the third could be any of 5+4 or 9..3C2*9C1=3*9=27 Total ways = (3+5+4)C3=12C3=12*11*10/3*2=220
Prob=27/220
Show more
Dear Sir
Thank you so much.
Can you also try and get the same result using my approach?
What can be the quickest approach to this question:-
A bag contains 3 Blue 5 Green 4 yellow balls 3 balls are drawn at random without replacement. What is the probability that exactly 2 of the 3 balls will be Blue.?
The approach i used is very lengthy and I think not correct or advisable
I used 1-(Prob of All Blue + Prob of no Blue + Prob of One Green)
Thanks
Posted from my mobile device
The correct way would be Ways in which 2 are blue 3C2 and the third could be any of 5+4 or 9..3C2*9C1=3*9=27 Total ways = (3+5+4)C3=12C3=12*11*10/3*2=220
Prob=27/220
Dear Sir
Thank you so much.
Can you also try and get the same result using my approach?
Thanks
Show more
So probability of all blue =3C3/12C3=1/220 Probability of one blue = 3C1*9C2/12C3=3*36/220=108/220 Probability of none blue =9C3/12C3=84/220 Total ways =1-(1/220+108/220+84/220)=1-(1+108+84)/220=1-193/220=27/220
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.