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Question
A bag contains 2 yellow, 3 green and 2 blue balls. Two balls are drawn at random. What is the probability that none of the balls drawn is blue?
A bag contains 2 yellow, 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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Solution
The correct option is B 10/21
Total number of balls = 2 + 3 + 2 = 7
Let S be the sample space.
n(S) = Total number of ways of drawing 2 balls out of 7 = 7C2
Let E = Event of drawing 2 balls , none of them is blue.
n(E) = Number of ways of drawing 2 balls , none of them is blue = Number of ways of drawing 2 balls from the total 5 (=7-2) balls = 5C2
(∵ There are two blue balls in the total 7 balls. Total number of non-blue balls = 7 - 2 = 5)
P(E)= n(E)/n(S)= 5C2 / 7C2 =
(5×42×1)/(7×62×1)=5×47×6=10/21
Total number of balls = 2 + 3 + 2 = 7
Let S be the sample space.
n(S) = Total number of ways of drawing 2 balls out of 7 = 7C2
Let E = Event of drawing 2 balls , none of them is blue.
n(E) = Number of ways of drawing 2 balls , none of them is blue = Number of ways of drawing 2 balls from the total 5 (=7-2) balls = 5C2
(∵ There are two blue balls in the total 7 balls. Total number of non-blue balls = 7 - 2 = 5)
P(E)= n(E)/n(S)= 5C2 / 7C2 =
(5×42×1)/(7×62×1)=5×47×6=10/21
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