In an examination, 20 questions of true-false type are asked. Suppose a student tosses a fair coin to determine his answer to each question. If the coin falls heads, he answers ‘true’; if it falls tails, he answers ‘false’. Find the probability that he answers at least 12 questions correctly.
In an examination, 20 questions of true-false type are asked. Suppose a student tosses a fair coin to determine his answer to each question. If the coin falls heads, he answers ‘true’; if it falls tails, he answers ‘false’. Find the probability that he answers at least 12 questions correctly.

Solution:

Assume that x is the number of correctly answered questions out of a total of twenty.

Since the ‘head’ of the coin represents the correct answer and the ‘tail’ represents the incorrect answer. Bernoulli trails are thus formed by repeated tosses or correctly answered questions.

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

Here, it can be clearly observed that x has binomial distribution, where n=20 and p=

Thus, \mathrm{P}(\mathrm{X}=\mathrm{x})={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{x}} \mathrm{q}^{n-\mathrm{x}} \mathrm{p}^{\mathrm{x}}, where \mathrm{x}=0,1,2 \ldots \mathrm{n}

={ }^{20} \mathrm{C}_{\mathrm{x}}\left(\frac{1}{2}\right)^{20-\mathrm{x}}\left(\frac{1}{2}\right)^{\mathrm{x}}

={ }^{20} \mathrm{C}_{\mathrm{x}}\left(\frac{1}{2}\right)^{20}

Probability of at least 12 questions answered correctly =\mathrm{p}(\mathrm{X} \geq 12)

    \[=P(X=12)+P(X=13)+\ldots+P(X=20)\]

={ }^{20} \mathrm{C}<em>{12}\left(\frac{1}{2}\right)^{20}+{ }^{20} \mathrm{C}</em>{13}\left(\frac{1}{2}\right)^{20}+\cdots+{ }^{20} \mathrm{C}_{20}\left(\frac{1}{2}\right)^{20}

=\left(\frac{1}{20}\right)^{20}\left({ }^{20} \mathrm{C}<em>{12}+{ }^{20} \mathrm{C}</em>{13}+\cdots+{ }^{20} \mathrm{C}_{20}\right)