Question
In a multiple-choice question, there are five alternative answers, of which one or more than one is correct. A candidate will get marks on the question if he ticks all the correct answers. If he decides to tick answers at random, how many minimum chances should he be allowed so that the probability of his getting marks on the question exceeds $\frac{1}{8}$.
Options
Solution
Sol. [Key Concept: Tick at random among $2^{5} - 1$ option-subsets; $P( \leq \ n\ tries) = n/31 > 1/8 \Rightarrow \ n \geq 4$.]
Let he takes ' $n$ ' chances
$$P = 1 - P(\text{~}\text{none correct}\text{~})$$
[ ∵ Total ways to answer $= 2^{5} - 1 = 31$ ]
$$\begin{matrix} & \ = 1 - \frac{\ ^{30}C_{n}}{31} = 1 - \frac{31 - n}{31} = \frac{n}{31} \\ \Rightarrow \ \frac{n}{31} > \frac{1}{8} & \ \Rightarrow n \geq 4 \end{matrix}$$
Alternate : $\rightarrow P$ (getting correct in $r^{\text{th}\text{~}}$ trial) $= \frac{\ ^{30}C_{r - 1}}{\ ^{31}C_{r - 1}} \times \frac{1}{32 - r} = \frac{1}{31}$
$\therefore P$ (correct in $1^{\text{st}\text{~}}$ trial or $2^{\text{nd}\text{~}}$ trial or ......or $n^{\text{th}\text{~}}$ trial)
$$= \frac{1}{31} + \frac{1}{31} + \frac{1}{31} + \ldots\ldots + \frac{1}{31} = \frac{n}{31}$$
[minimum value of $n = 4\rbrack$
Create a free account to view solution
View Solution Free