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Question
A bag contains 5 red balls, 6 white balls, 7 green balls, 8 black balls. One ball is drawn at random from the bag. Find the probability that the ball drawn is neither white or black
Solution
Sample space (S) = 5 + 6 + 7 + 8
n(S) = 26
Let A be the event of getting a white ball
n(A) = 6
P(A) = `("n"("A"))/("n"("S")) = 6/26`
Let B be the event of getting a black ball
n(B) = 8
P(B) = `("n"("B"))/("n"("S")) = 8/26`
P(A ∪ B) = P(A) + P(B)
= `6/26 + 8/26`
= `14/26`
Probability of neither white or black P(A ∪ B)’ = 1 – P(A ∪ B)
= `1 - 14/26`
= `(26 - 14)/26`
= `12/26`
= `6/13`
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