If 3 numbers are chosen from the first 15 nat | vedclass

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(D) The total number of ways to choose $3$ numbers from $15$ is $^{15}C_3 = \frac{15 	imes 14 	imes 13}{3 	imes 2 	imes 1} = 455$. For $3$ numbers

Vedclass · Accepted Answer

$\frac{7}{65}$

Vedclass · Answer

(D) The total number of ways to choose $3$ numbers from $15$ is $^{15}C_3 = \frac{15 	imes 14 	imes 13}{3 	imes 2 	imes 1} = 455$. For $3$ numbers $(a, b, c)$ to be in arithmetic progression,$a+c = 2b$,which implies $a+c$ must be even. This happens if both $a$ and $c$ are even or both are odd. Number of even integers in ${1, 2, \dots, 15}$ is $7$ and odd integers is $8$. Number of ways to choose $2$ even numbers is $^7C_2 = 21$. Number of ways to choose $2$ odd numbers is $^8C_2 = 28$. Total favorable outcomes = $21 + 28 = 49$. Probability = $\frac{49}{455} = \frac{7}{65}$.

If $3$ numbers are chosen from the first $15$ natural numbers,what is the probability that the numbers are in arithmetic progression?

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