Question

In a room filled with 7 people, 4 people have exactly 1 sibling in the room and 3 people have exactly 2 siblings in the room. If two individuals are selected from the room at random, what is the probability that those two individuals are NOT siblings?
A. 5/21 B. 3/7 C. 4/7 D. 5/7 E. 16/21

Answer 1

Answer:

Option E

$\frac{16}{21}$

Step-by-step explanation:

Number of ways of selecting two people from a group of seven people is

$C^2_{7}$

$\frac{7 * 6* 5!}{2*1* 5!} \\= 21$

Number of ways by which two people selected in a room come out to be siblings are -

$C^2_{2}$+$C^2_{3}$+ $C^2_{2}$

$= \frac{2*1}{2*1} + \frac{3*2*1}{2*1} + \frac{2*1}{2*1}\\= 1+ 3+1\\= 5$

Number of ways by which two people selected in a room are not siblings

$= 1-$Probability of selecting two siblings

$= 1-\frac{5}{21} \\= \frac{16}{21}$

Hence, option E is correct