A) \[\frac{1}{3}\]
B) \[\frac{2}{8}\]
C) \[\frac{2}{5}\]
D) \[\frac{2}{3}\]
Correct Answer: A
Solution :
Random drawing of blocks ensures equally likely outcomes. The total number of blocks representing A = 2 Total number of possible outcomes \[=6P(A)=\] \[\frac{\text{Number}\,\text{of}\,\text{favourable}\,\text{outcomes}}{\text{Total}\,\text{number}\,\text{of}\,\text{possible}\,\text{outcomes}}\] \[\text{=}\frac{2}{6}=\frac{1}{3}\]You need to login to perform this action.
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