Question

A buys 3 tickets of a lottery in which total number of tickets is 12 out of which 3 carry prizes on them, A's chance of getting a prize is

• A. $\frac{37}{52}$
• B. $\frac{34}{55}$
• C. $\frac{41}{55}$
• D. None of these

Answer 1

The correct option is B $\frac{34}{55}$

A can get prize if 1, 2 or all 3 tickets that he has bought has prizes.
hence $P\left(Awin\right)=1-P\left(noneofthembeingprizetickets\right)=1-\frac{\left(\begin{array}{c}9\\ 3\end{array}\right)}{\left(\begin{array}{c}12\\ 3\end{array}\right)}=\frac{136}{220}=\frac{34}{55}$