In a box, there are 8 red, 7 blue, and 6 green balls. One ball is picked randomly. The probability that it is neither red nor green is
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- $\dfrac{1}{3}$
- $\dfrac{3}{4}$
- $\dfrac{7}{19}$
- $\dfrac{8}{21}$
The Correct Option is A
Solution and Explanation
Balls that are neither red nor green = blue balls = 7
So required probability = $\dfrac{7}{21} = \dfrac{1}{3}$
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Let \( A_1 \): People with good health,
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