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Question
Mark the correct alternative in the following question:
The probability of guessing correctly at least 8 out of 10 answers of a true false type examination is
Options
\[\frac{7}{64}\]
\[\frac{7}{128}\]
\[\frac{45}{1024} \]
\[\frac{7}{41}\]
Solution
\[\text{ We have,} \]
\[p = \text{ probabiltiy of guessing the answer of a true false correctly } = \frac{1}{2} \text{ and } \]
\[q = \text{ probabiltiy of guessing the answer of a true false incorrectly } = 1 - p = 1 - \frac{1}{2} = \frac{1}{2}\]
\[\text{ Let X denote a success of guessing the answer correctly . Then, } \]
\[\text{ X follows the binomial distribution with parameters n = 10 and } p = \frac{1}{2}\]
\[ \therefore P\left( X = r \right) = ^{10}{}{C}_r p^r q^\left( 10 - r \right) = ^{10}{}{C}_r \left( \frac{1}{2} \right)^r \left( \frac{1}{2} \right)^\left( 10 - r \right) = ^{10}{}{C}_r \left( \frac{1}{2} \right)^{10} = \frac{^{10}{}{C}_r}{2^{10}}\]
\[\text{ Now } , \]
\[\text{ Required probability } = P\left( X \geq 8 \right)\]
\[ = P\left( X = 8 \right) + P\left( X = 9 \right) + P\left( X = 10 \right)\]
\[ = \frac{^{10}{}{C}_8}{2^{10}} + \frac{^{10}{}{C}_9}{2^{10}} + \frac{^{10}{}{C}_{10}}{2^{10}}\]
\[ = \frac{45 + 10 + 1}{2^{10}}\]
\[ = \frac{56}{1024}\]
\[ = \frac{7}{128}\]
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