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A bag contains 2 red, 3 green and 2 blue balls. Two balls are drawn at random. What is the probability that none of the balls drawn is blue?

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((5!*5!)/(3!*7!))=10/21=0.476190
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10/21
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Deepak 2C2/7C2 = 1/21 ? Probability=Possible out come / Total out come. 5C2/7C2 = 10/21 5C2 -> 7C2 ->
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5/7
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The Complete Cover

1-(10/21)=11/21
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Solution: Total Number of balls = 2+2+3= 7 Number of ways of drawing 2 balls out of 7 = 7C2 Total Number of balls other than Blue = 2 red + 3 green = 5 Number of ways of drawing 2 balls( other than blue) out of 7 = 7C2 Probability of drawing balls other than blue = 5C2/7C2 =10/21
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10/21
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Total balls is 7. Number of ways of drawing 2 out of this 7 is 21. Event of drawing 2 balls, none of which is blue is 10. Therefore, answer is 10/21
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5/7 * 2/3
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((5!*5!)/(3!*7!))=10/21=0.476190
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If the sum of two numbers & their LCM is 79.Then how many possible set of numbers can satisfy above equation.
Hi Sathya - there are a set of solutions - Let the numbers be A and B and their LCM be L and HCF be H. Thus, A and B are multiples of H. Let A = H*p; B = H*q. Thus, L = H*p*q; where p and q are co-prime. Hence,...
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If increasing 20 by P percentage gives the same result as decreasing 60 by P percentage,What is P percentage of 70?
Increasing 20 by P% => x=20+20.P/100 Decreasing 60 by P% => y=60-60.P/100 Acoording to the question, x=y => 20+20.P/100=60-60.P/100 Solving we get P=50 So, P% of 70=50% of 70=35 So, answer is 35
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