CBSE 2023 · Region 2 · Set 1 · Q30 · 3 marks
Two balls are drawn at random one by one with replacement from an urn containing equal number of red balls and green balls. Find the probability distribution of number of red balls. Also, find the mean of the random variable.A and B throw a die alternately till one of them gets a ' $\displaystyle 6$ ' and wins the game. Find their respective probabilities of wining, if A starts the game first.
Two balls are drawn at random one by one with replacement from an urn containing equal number of red balls and green balls. Find the probability distribution of number of red balls. Also, find the mean of the random variable.
A and B throw a die alternately till one of them gets a ' $\displaystyle 6$ ' and wins the game. Find their respective probabilities of wining, if A starts the game first.
Marking-scheme solution
(a)
X: Number of red balls out of the two balls drawn
$$\begin{gathered}
\therefore \text { Mean }=$\displaystyle 0$ \cdot \frac{1}{4}+$\displaystyle 1$ \cdot \frac{1}{2}+$\displaystyle 2$ \cdot \frac{1}{4}=$\displaystyle 1$
\text { Or }
\end{gathered}
$$(b) $\displaystyle \mathrm{P}($ getting a six $\displaystyle )=\frac{1}{6} ; \mathrm{P}($ not getting a six $\displaystyle )=\frac{5}{6}$
$\displaystyle \mathrm{P}($ A wins $\displaystyle )=\frac{1}{6}+\left(\frac{5}{6}\right)^{2} \cdot \frac{1}{6}+\left(\frac{5}{6}\right)^{4} \cdot \frac{1}{6}+\ldots=\frac{\dfrac{1}{6}}{1-\dfrac{25}{36}}=\frac{6}{11}$
$$\mathrm{P}(B \text { wins })=$\displaystyle 1$-\mathrm{P}(A \text { wins })=$\displaystyle 1$-\frac{6}{11}=\frac{5}{11}
ProbabilityRandom Variable and its Probability DistributionApplyshort_answermedium
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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.