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Home » On a multiple choice examination with three possible answers for each of the five questions, what is the probability that a candidate would get four or more correct answers just by guessing?
On a multiple choice examination with three possible answers for each of the five questions, what is the probability that a candidate would get four or more correct answers just by guessing?
CBSE | Maths | NCERT | Probability
On a multiple choice examination with three possible answers for each of the five questions, what is the probability that a candidate would get four or more correct answers just by guessing?

Solution:

Bernoulli trials are the repeated correct answer guessing from the given multiple choice questions in this question.

Assume that X represents the number of correct answers in the multiple choice set obtained by guessing.
Now, probability of getting a correct answer, p=1 / 3

Thus, q=1-p=1-1 / 3=2 / 3

Clearly, we have \mathrm{X} is a binomial distribution where \mathrm{n}=5 and \mathrm{P}=1 / 3

\therefore P(X=x)={ }^{n} C_{x} q^{n-x} p^{x}

={ }^{5} \mathrm{C}_{\mathrm{x}}\left(\frac{2}{3}\right)^{5^{-\mathrm{x}}} \cdot\left(\frac{1}{3}\right)^{\mathrm{x}}

Hence, probability of guessing more than 4 correct answer =\mathrm{P}(\mathrm{X} \geq 4)

=\mathrm{P}(\mathrm{X}=4)+\mathrm{P}(\mathrm{X}=5)

={ }^{5} \mathrm{C}{4}\left(\frac{2}{3}\right) \cdot\left(\frac{1}{3}\right)^{4}+{ }^{5} \mathrm{C}{5}\left(\frac{1}{3}\right)^{5}

=5 \cdot \frac{2}{3} \cdot \frac{1}{81}+1 \cdot \frac{1}{243}

=\frac{11}{243}

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