ProbabilityHardBloom L3
Question
3 distinct numbers are chosen from the set of first 15 natural numbers, then the probability that the numbers are in A.P. is equal to
Options
A.$\frac{51}{455}$
B.$\frac{7}{65}$
C.$\frac{10}{91}$
D.$\frac{53}{455}$
Solution
Sol. [Key Concept: $3$ distinct from the first 15 naturals under the condition.]
$$P = \frac{\ ^{8}C_{2} + \ ^{7}C_{2}}{\ ^{15}C_{3}} = \frac{7}{65}$$
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