CBSE 2023 · Region 3 · Set 1 · Q31 · 3 marks
A pair of dice is thrown simultaneously. If X denotes the absolute difference of numbers obtained on the pair of dice, then find the probability distribution of X.There are two coins. One of them is a biased coin such that P (head) : P (tail) is $\displaystyle 1: 3$ and the other coin is a fair coin. A coin is selected at random and tossed once. If the coin showed head, then find the probability that it is a biased coin.
A pair of dice is thrown simultaneously. If X denotes the absolute difference of numbers obtained on the pair of dice, then find the probability distribution of X.
There are two coins. One of them is a biased coin such that P (head) : P (tail) is $\displaystyle 1: 3$ and the other coin is a fair coin. A coin is selected at random and tossed once. If the coin showed head, then find the probability that it is a biased coin.
Marking-scheme solution
\multicolumn{5}{|l|}{(a)
$\displaystyle \mathrm{P}(\mathrm{X})$ & $\displaystyle \frac{6}{36}$ & $\displaystyle \frac{10}{36}$ & $\displaystyle \frac{8}{36}$ & $\displaystyle \frac{6}{36}$ & $\displaystyle \frac{4}{36}$ & $\displaystyle \frac{2}{36}$}
\multicolumn{5}{|c|}{
(b)
$\displaystyle \mathrm{E}_{1}=$ Biased coin is selected $\displaystyle \Rightarrow \mathrm{P}\left(\mathrm{E}_{1}\right)=\frac{1}{2}$
$$\mathrm{E}_{$\displaystyle 2$}=\text { Fair coin is selected } \Rightarrow \mathrm{P}\left(\mathrm{E}_{$\displaystyle 2$}\right)=\frac{1}{2}
$$
A = Head appeared on tossing a selected coin .
$$\mathrm{P}\left(\frac{A}{\mathrm{E}_{$\displaystyle 1$}}\right)=\frac{1}{4}, \mathrm{P}\left(\frac{A}{\mathrm{E}_{$\displaystyle 2$}}\right)=\frac{1}{2}
$$By Bayes' Theorem $\displaystyle \mathrm{P}\left(\frac{\mathrm{E}_{1}}{A}\right)=\frac{\mathrm{P}\left(\mathrm{E}_{1}\right) \mathrm{P}\left(\dfrac{A}{\mathrm{E}_{1}}\right)}{\mathrm{P}\left(\mathrm{E}_{1}\right) \mathrm{P}\left(\dfrac{A}{\mathrm{E}_{1}}\right)+\mathrm{P}\left(\mathrm{E}_{2}\right) \mathrm{P}\left(\dfrac{A}{\mathrm{E}_{2}}\right)}$
$$\begin{gathered}
=\frac{\dfrac{1}{2} \cdot \dfrac{1}{4}}{\dfrac{1}{2} \cdot \dfrac{1}{4}+\dfrac{1}{2} \cdot \dfrac{1}{2}}
=\frac{1}{3}
\end{gathered}
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CBSE Class 12 Mathematics past-paper question from the 2023board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.