Probability
Balls in Boxes
MMTS_Full_Test_19
Grade 12
Question:
Three distinct numbers are selected from first 100 natural numbers. The probability that all the three numbers are divisible by both 2 and 3 is
$\dfrac{4}{1155}$
$\dfrac{4}{25}$
$\dfrac{4}{35}$
$\dfrac{4}{55}$
Step-by-Step Solution
Key Concept: Numbers divisible by both 2 and 3 means divisible by 6; count multiples of 6 up to 100
16 multiples of 6. $P=\frac{\binom{16}{3}}{\binom{100}{3}}=\frac{560}{161700}=\frac{4}{1155}$.
Correct Answer: 1