GO Classes 2024 | Weekly Quiz 8 | Conditional Probability | Question: 13

Answer 1

$Box$ contains $3$ coins; So, choosing randomly one coin will be equally likely.

$\therefore$ Probability of choosing $1$ coin = $P(C_{1}) = P(C_{2}) = P(C_{3}) = \frac{1}{3}$

Given that, there are $2$ regular coins & $1$ fake two-headed coin $(P(H) = 1)$.

So, for $2$ regular coins the occurrence of $Head$ and $Tail$ will be equal individually.

$\therefore$ $P(H|C_{1}) = P(T|C_{1}) = \frac{1}{2}$ & $P(H|C_{2}) = P(T|C_{2}) = \frac{1}{2}$

$\therefore$ Probability that it lands up heads = $P(H|C_{1}).P(C_{1}) + P(H|C_{2}).P(C_{2}) + P(C_{3}).P(H|C_{3})$

 = $\frac{1}{2} \times \frac{1}{3} + \frac{1}{2} \times \frac{1}{3} + \frac{1}{3} \times 1 = \frac{1}{6} + \frac{1}{6} + \frac{1}{3} = \frac{2}{3}$

$Ans: B. 2/3$   

Answer 2

This is another typical problem for which the law of total probability is useful. Let $\mathrm{C}_1$ be the event that you choose a regular coin, and let $\mathrm{C}_2$ be the event that you choose the two-headed coin. Note that $\mathrm{C}_1$ and $\mathrm{C}_2$ form a partition of the sample space. We already know that
$$
\begin{gathered}
\mathrm{P}\left(\mathrm{H} \mid \mathrm{C}_1\right)=0.5, \\
\mathrm{P}\left(\mathrm{H} \mid \mathrm{C}_2\right)=1 .
\end{gathered}
$$
Thus, we can use the law of total probability to write
$$
\begin{aligned}
\mathrm{P}(\mathrm{H}) & =\mathrm{P}\left(\mathrm{H} \mid \mathrm{C}_1\right) \mathrm{P}\left(\mathrm{C}_1\right)+\mathrm{P}\left(\mathrm{H} \mid \mathrm{C}_2\right) \mathrm{P}\left(\mathrm{C}_2\right) \\
& =\frac{1}{2} \cdot \frac{2}{3}+1 \cdot \frac{1}{3} \\
& =\frac{2}{3}.
\end{aligned}
$$

Answer 3

$P(H) = 1/3 + 1/2*1/3 + 1/2*1/3 = 1/3 +1/3 = 2/3$