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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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