A multiple choice examination has 5 questions. Each question has three alternative answers of which exactly…

A multiple choice examination has 5 questions. Each question has three alternative answers of which exactly one is correct. The probability that a student will get 4 or more correct answers just by guessing is
  1. $\frac{17}{243}$
  2. $\frac{13}{243}$
  3. $\frac{11}{243}$
  4. $\frac{10}{243}$

Solution

Probability of getting correct answer $(p)=\frac{1}{3}$ $\therefore \quad q=1-\frac{1}{3}=\frac{2}{3}$
Also, $\mathrm{n}=5$ $\therefore \quad$ Required probability $\begin{aligned} & =\mathrm{P}(\mathrm{X} \geq 4) \\ & =\mathrm{P}(\mathrm{X}=4)+\mathrm{P}(\mathrm{X}=5) \\ & ={ }^5 \mathrm{C}_4\left(\frac{1}{3}\right)^4\left(\frac{2}{3}\right)^1+{ }^5 \mathrm{C}_5\left(\frac{1}{3}\right)^5\left(\frac{2}{3}\right)^0 \\ & =5\left(\frac{1}{3}\right)^4\left(\frac{2}{3}\right)+1\left(\frac{1}{3}\right)^5 \\ & =\left(\frac{1}{3}\right)^4\left(\frac{10}{3}+\frac{1}{3}\right)=\frac{11}{3^5}=\frac{11}{243} \end{aligned}$

Asked in: MHT CET 2024 (15 May Shift 1)

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