A student writes an examination which contains eight true or false questions. If he answers six or more…

A student writes an examination which contains eight true or false questions. If he answers six or more questions correctly, he passes the examination. If the student answers all the questions, then the probability that he fails in the examination is
  1. $\frac{37}{256}$
  2. $\frac{19}{256}$
  3. $\frac{119}{256}$
  4. $\frac{219}{256}$

Solution

Given, 8 True/False question So, $\mathrm{P}($ Answer correctly $)=\frac{1}{2}$ $\mathrm{P}($ Answer incorrectly $)=\frac{1}{2}$ $\begin{aligned} & \text { P (Student pass) } \\ & ={ }^8 \mathrm{C}_6\left(\frac{1}{2}\right)^6\left(\frac{1}{2}\right)^2+{ }^8 \mathrm{C}_7\left(\frac{1}{2}\right)^7\left(\frac{1}{2}\right)^1+{ }^8 \mathrm{C}_8\left(\frac{1}{2}\right)^8\end{aligned}$ $=\frac{\left(\frac{8 \times 7}{2}+8+1\right)}{2^8}=\frac{37}{256}$ Now, $P($ Student fails $)=1-\frac{37}{256}=\frac{219}{256}$

Asked in: AP EAMCET 2024 (18 May Shift 1)

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