A bag contains 2 red, 3 green and 2 blue balls. Two balls are drawn without replacement. What is the probability that neither ball drawn is blue?

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Multiple Choice

A bag contains 2 red, 3 green and 2 blue balls. Two balls are drawn without replacement. What is the probability that neither ball drawn is blue?

Explanation:
You're testing the probability that both drawn balls are not blue. There are five non-blue balls (red or green) among seven total. The chance both draws avoid blue can be found by counting: choose two from the five non-blue is C(5,2) = 10, while total two-ball draws from seven is C(7,2) = 21. So the probability is 10/21. An equivalent sequential view is: first draw is non-blue (5/7), then the second is non-blue given the first was non-blue (4/6 = 2/3). Multiply: (5/7) * (2/3) = 10/21. Therefore, the probability is 10/21.

You're testing the probability that both drawn balls are not blue. There are five non-blue balls (red or green) among seven total. The chance both draws avoid blue can be found by counting: choose two from the five non-blue is C(5,2) = 10, while total two-ball draws from seven is C(7,2) = 21. So the probability is 10/21. An equivalent sequential view is: first draw is non-blue (5/7), then the second is non-blue given the first was non-blue (4/6 = 2/3). Multiply: (5/7) * (2/3) = 10/21. Therefore, the probability is 10/21.

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