A bag contains 6 white,5 black,and 4 red ball | vedclass

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(B) Total number of balls in the bag $= 6 + 5 + 4 = 15$.Let $A$ be the event of getting a white ball and $B$ be the event of getting a black ball.The

Vedclass · Accepted Answer

$11/15$

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(B) Total number of balls in the bag $= 6 + 5 + 4 = 15$.Let $A$ be the event of getting a white ball and $B$ be the event of getting a black ball.The number of white balls is $6$,so $P(A) = \frac{6}{15}$.The number of black balls is $5$,so $P(B) = \frac{5}{15}$.Since the events $A$ and $B$ are mutually exclusive,the probability of getting a white or a black ball is $P(A \cup B) = P(A) + P(B)$.$P(A \cup B) = \frac{6}{15} + \frac{5}{15} = \frac{11}{15}$.

$A$ bag contains $6$ white,$5$ black,and $4$ red balls. Find the probability of getting a white or a black ball when one ball is drawn from the bag.

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