Question

A set \(A\) contains first 30 natural numbers. If 3 numbers are chosen at random from the set \(A\), the probability that their product is an odd integer is

• A. \(\frac{45}{116}\)
• B. \(\frac{103}{116}\)
• C. \(\frac{71}{116}\)
• D. \(\frac{13}{116}\)

Hint:

A product is odd only if every factor is odd. Therefore, count selections entirely from the odd numbers.

Answer 1

The Correct Option is D

Concept: The product of selected numbers will be odd only when all the chosen numbers are odd.

Step 1: Count odd numbers in the set. The first 30 natural numbers are \[ 1,2,3,\ldots,30 \] Among them, \[ 15 \] are odd and \[ 15 \] are even.

Step 2: Find the total number of ways of choosing 3 numbers. \[ {}^{30}C_3 = \frac{30\times29\times28}{3\times2\times1} = 4060 \]

Step 3: Find the favourable outcomes. To obtain an odd product, all three chosen numbers must be odd. Number of favourable selections \[ {}^{15}C_3 = \frac{15\times14\times13}{3\times2\times1} = 455 \]

Step 4: Compute the probability. \[ P = \frac{{}^{15}C_3}{{}^{30}C_3} = \frac{455}{4060} \] \[ = \frac{91}{812} = \frac{13}{116} \] Therefore, \[ P=\frac{13}{116} \] \[ \boxed{\frac{13}{116}} \]