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

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

Class 11 Mathematics Probability Advanced Use of permutations and combinations in probability

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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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A
$\frac{496}{1001}$
B
$\frac{505}{1001}$
C
$\frac{490}{1001}$
D
$\frac{504}{1001}$

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