Six distinct numbers are chosen from the first 10 natural numbers. The probability that 6 is the third largest of those chosen numbers, is

Question

Six distinct numbers are chosen from the first 10 natural numbers. The probability that 6 is the third largest of those chosen numbers, is (A) (2/7) (B) (5/21) (C) (10/63) (D) (16/63)

Tardigrade Admin · Accepted Answer

Total cases = 10C6 Favourable cases = 4C2⋅ 1C1⋅ 5C3 Required probability = (4C2 ⋅ 1 C1 ⋅ 5 C3/ 10C6)=(2/7)

Q. Six distinct numbers are chosen from the first $10$ natural numbers. The probability that $6$ is the third largest of those chosen numbers, is

$\frac{2}{7}$

$\frac{5}{21}$

$\frac{10}{63}$

$\frac{16}{63}$

Total cases = $_^{10} C_{6}$
Favourable cases = $^{4} C_{2} \cdot^{1} C_{1} \cdot^{5} C_{3}$
Required probability = $\frac{^{4} C _{2} \cdot ^{1} C _{1} \cdot ^{5} C _{3}}{_ ^{10} C _{6}} = \frac{2}{7}$

Q. Six distinct numbers are chosen from the first 10 natural numbers. The probability that 6 is the third largest of those chosen numbers, is

72​

215​

6310​

6316​