MHT CET · Maths · Probability
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
- A \(\frac{10}{3^5}\)
- B \(\frac{17}{3^5}\)
- C \(\frac{13}{3^5}\)
- D \(\frac{11}{3^5}\)
Answer & Solution
Correct Answer
(D) \(\frac{11}{3^5}\)
Step-by-step Solution
Detailed explanation
Probability of guessing correct answer be
\(p=\frac{1}{3}\)
\(\therefore \mathrm{q}=\frac{2}{3}\)
Let random variable X denotes the number of correct answers.
\(\therefore X \sim B\left(5, \frac{1}{3}\right) \)
\( \therefore \text { Required probability }=\mathrm{P}(\mathrm{X} \geq 4) \)
\( =\mathrm{P}(\mathrm{X}=4)+\mathrm{P}(\mathrm{X}=5) \)
\( ={ }^5 \mathrm{C}_4\left(\frac{1}{3}\right)^4\left(\frac{2}{3}\right)+{ }^5 \mathrm{C}_5\left(\frac{1}{3}\right)^5 \)
\( =5 \times \frac{2}{3^5}+\frac{1}{3^5} \)
\( =\frac{11}{3^5}\)
\(p=\frac{1}{3}\)
\(\therefore \mathrm{q}=\frac{2}{3}\)
Let random variable X denotes the number of correct answers.
\(\therefore X \sim B\left(5, \frac{1}{3}\right) \)
\( \therefore \text { Required probability }=\mathrm{P}(\mathrm{X} \geq 4) \)
\( =\mathrm{P}(\mathrm{X}=4)+\mathrm{P}(\mathrm{X}=5) \)
\( ={ }^5 \mathrm{C}_4\left(\frac{1}{3}\right)^4\left(\frac{2}{3}\right)+{ }^5 \mathrm{C}_5\left(\frac{1}{3}\right)^5 \)
\( =5 \times \frac{2}{3^5}+\frac{1}{3^5} \)
\( =\frac{11}{3^5}\)
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