MHT CET · Maths · Probability
There are 6 positive and 8 negative numbers. From these four numbers are chosen at random and multiplied. Then the probability, that the product is a negative number, is
- A \(\frac {496}{1001}\)
- B \(\frac {505}{1001}\)
- C \(\frac {490}{1001}\)
- D \(\frac {504}{1001}\)
Answer & Solution
Correct Answer
(A) \(\frac {496}{1001}\)
Step-by-step Solution
Detailed explanation
Total number of numbers \(=8+6=14\) 4 numbers can be chosen out of 14 numbers in \({ }^{14} \mathrm{C}_4\) ways
The product of 4 numbers will be negative, if i. one is negative and three are positive
OR
ii. three are negative and one is positive.
\(\therefore \text {Required probability} =\frac{\left({ }^8 \mathrm{C}_1 \times{ }^6 \mathrm{C}_3\right)+\left({ }^8 \mathrm{C}_3 \times{ }^6 \mathrm{C}_1\right)}{{ }^{14} \mathrm{C}_4}\)
\(=\frac{(8 \times 20)+(56 \times 6)}{1001}\)
\(=\frac{160+336}{1001}=\frac{496}{1001}\)
The product of 4 numbers will be negative, if i. one is negative and three are positive
OR
ii. three are negative and one is positive.
\(\therefore \text {Required probability} =\frac{\left({ }^8 \mathrm{C}_1 \times{ }^6 \mathrm{C}_3\right)+\left({ }^8 \mathrm{C}_3 \times{ }^6 \mathrm{C}_1\right)}{{ }^{14} \mathrm{C}_4}\)
\(=\frac{(8 \times 20)+(56 \times 6)}{1001}\)
\(=\frac{160+336}{1001}=\frac{496}{1001}\)
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