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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?
Options:
A .  $MF#%\dfrac{1}{2}$MF#%
B .  $MF#%\dfrac{10}{21}$MF#%
C .  $MF#%\dfrac{9}{11}$MF#%
D .  $MF#%\dfrac{7}{11}$MF#%
Answer: Option B

Answer : Option B

Explanation :

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)

$MF#%\text{P(E) = }\dfrac{\text{n(E)}}{\text{n(S)}}= \dfrac{5_{C_2}}{7_{C_2}}
= \dfrac{\left(\dfrac{5 \times 4 }{2 \times 1} \right )}{\left(\dfrac{7 \times 6 }{2 \times 1} \right )}= \dfrac{5 \times 4}{7 \times 6} = \dfrac{10}{21}$MF#%



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