A bag contains 5 white,7 red,and 8 black ball | vedclass

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Question

(A) Total number of balls $= 5 + 7 + 8 = 20$.We need to draw $4$ white balls one by one without replacement.The probability of drawing the first white

Vedclass · Accepted Answer

$\frac{1}{969}$

Vedclass · Answer

(A) Total number of balls $= 5 + 7 + 8 = 20$.We need to draw $4$ white balls one by one without replacement.The probability of drawing the first white ball is $\frac{5}{20}$.Since the ball is not replaced,the number of white balls becomes $4$ and the total number of balls becomes $19$. The probability of drawing the second white ball is $\frac{4}{19}$.Similarly,the probability of drawing the third white ball is $\frac{3}{18}$,and the probability of drawing the fourth white ball is $\frac{2}{17}$.Therefore,the required probability is $\frac{5}{20} 	imes \frac{4}{19} 	imes \frac{3}{18} 	imes \frac{2}{17} = \frac{1}{4} 	imes \frac{4}{19} 	imes \frac{1}{6} 	imes \frac{2}{17} = \frac{1}{19} 	imes \frac{1}{3} 	imes \frac{1}{17} = \frac{1}{969}$.

$A$ bag contains $5$ white,$7$ red,and $8$ black balls. If four balls are drawn one by one without replacement,what is the probability that all are white?

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