Question

Two numbers are selected at random from $1,2,3,4,5,6$ without replacement. The probability that the minimum of the two numbers is less than $4$ is

• A. $\frac{4}{5}$
• B. $1-\frac{{}^{30}{C}_2}{{}^{100}{C}_2}$
• C. $\frac{{}^{31}{C}_2}{{}^{100}{C}_2}$
• D. $1-\frac{{}^{31}{C}_2}{{}^{100}{C}_2}$

Answer 1

The correct option is A $\frac{4}{5}$

Given set $S=1,2,3,4,5,6$

All the possible combinations are ${}^6{C}_2=15$.

Out of this, $12$ possible combinations where minimum of the two number is less than $44$ are

$(1,2),(1,3),(1,4),(1,5),(1,6),(2,3),(2,4),(2,5),(2,6),(3,4),(3,5),(3,6)$.

Hence the required probability $=\frac{12}{15}=\frac{4}{5}$