Question

Consider two sets $A=\{1,2,3,4,5\}$ and $B=\{1,2,3,\dots \dots ,9,10\}$.
If Tina randomly selects two distinct numbers from set A and Reena randomly selects one number from set B, the probability that Reena's number is greater than the sum of the two number chosen by Tina, is

• A. $\frac{2}{5}$
• B. $\frac{1}{2}$
• C. $\frac{3}{5}$
• D. $\frac{1}{5}$

Answer 1

Answer: (A)

For Tina, $n(S)={}^5{C}_2$
i.e. Two numbers selected are $12,13,14,15,23,24,25,34,35,45$ sum is $3,4,5,6,5,6,7,7,8,9$
Now for Reena, one number has to be selected so that sum is greater than the sum of the two numbers chosen by Tina
$\text{ Required probability }=\frac{1}{10}\left(\frac{7+6+5+4+5+4+3+3+2+1}{10}\right)$
$=\frac{40}{100}=\frac{4}{10}=\frac{2}{5}$