There are 6 positive and 8 negative numbers. | vedclass

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Question

(A) Total number of numbers $= 6 + 8 = 14$. Number of ways to choose $4$ numbers out of $14$ is $^{14}C_4 = \frac{14 	imes 13 	imes 12 	imes 11}{4

Vedclass · Accepted Answer

$\frac{496}{1001}$

Vedclass · Answer

(A) Total number of numbers $= 6 + 8 = 14$. Number of ways to choose $4$ numbers out of $14$ is $^{14}C_4 = \frac{14 	imes 13 	imes 12 	imes 11}{4 	imes 3 	imes 2 	imes 1} = 1001$. The product of $4$ numbers is negative if: $i$. One number is negative and three are positive: $^{8}C_1 	imes ^{6}C_3 = 8 	imes 20 = 160$. $ii$. Three numbers are negative and one is positive: $^{8}C_3 	imes ^{6}C_1 = 56 	imes 6 = 336$. Total favorable ways $= 160 + 336 = 496$. Required probability $= \frac{496}{1001}$.

There are $6$ positive and $8$ negative numbers. If four numbers are chosen at random and multiplied,the probability that the product is a negative number is

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