Question

Three distinct numbers are selected from 100 natural number. The probability that all the three numbers are divisible by 2

and 3 is

• A. $\frac { 4 } { 25 }$
• B. $\frac { 4 } { 35 }$
• C. $\frac { 4 } { 55 }$
• D. $\frac { 4 } { 1155 }$

Answer 1

Answer: (D)

$\frac { 4 } { 1155 }$

Answer 2

The numbers should be divisible by 6. Thus the number of favourable ways is ${ } ^ { 16 } C _ { 3 }$ (as there are 16 numbers in first 100 natural numbers, divisible by 6). Required probability is $\frac { { } ^ { 16 } C _ { 3 } } { { } ^ { 100 } C _ { 3 } } = \frac { 16 \times 15 \times 14 } { 100 \times 99 \times 98 } = \frac { 4 } { 1155 }$ .