Probability - Arithmetic Aptitude Questions and Answers

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In a box, there are 8 red, 7 blue and 6 green balls. One ball is picked up randomly. What is the probability that it is neither red nor green?

$\frac { 1 } { 3 }$

$\frac { 3 } {4 }$

$\frac { 7 } {19 }$

$\frac {8 } { 21 }$

$\frac {9 } { 21 }$

Answer : Option A

Explanation :

Total number of balls = (8 + 7 + 6) = 21.

Let E	= event that the ball drawn is neither red nor green
	= event that the ball drawn is blue.

n(E) = 7.

P(E) = $\frac { n(E) } { n(S)}$ = $\frac {7} { 21 }$ = $\frac { 2 } { 3 }$

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