Jamia Millia Islamia MCA Probability Distribution Previous Year Questions (PYQs)
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Three balls are drawn from a bag containing 2 red and 5 black balls.
If the random variable $X$ represents the number of red balls drawn,
then $X$ can take values……
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Solution
There are 2 red balls in total, so when drawing 3 balls,
possible red counts are 0 (no red), 1 (one red), or 2 (both reds).
$X$ can take values $\{0, 1, 2\}$.
100 identical coins, each with probability $p$ of showing heads, are tossed.
If $0 < p < 1$ and the probability of showing heads on 50 coins is equal to that of 51 coins, then the value of $p$ is:
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Solution
$\binom{100}{50}p^{50}(1-p)^{50} = \binom{100}{51}p^{51}(1-p)^{49}$
$\Rightarrow \dfrac{\binom{100}{50}}{\binom{100}{51}} \cdot \dfrac{1-p}{p} = 1$
$\dfrac{\binom{100}{50}}{\binom{100}{51}} = \dfrac{51}{50}$
$\Rightarrow \dfrac{1-p}{p} = \dfrac{50}{51}$
$\Rightarrow p = \dfrac{51}{101}$
The variance of first $50$ even natural numbers is
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Solution
Numbers are $2,4,\dots,100=2\cdot{1,\dots,50}$.
$\operatorname{Var}(1,\dots,50)=\dfrac{50^2-1}{12}=\dfrac{2499}{12}=\dfrac{833}{4}$.
Scaling by $2$: $\operatorname{Var}=4\cdot\dfrac{833}{4}=833$.
For two datasets each of size $5$, variances are $4$ and $5$ and means are $2$ and $4$ respectively.
The variance of the combined dataset is:
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Solution
Let $n_1 = n_2 = 5, ; \sigma_1^2 = 4, ; \sigma_2^2 = 5, ; \bar{x}_1 = 2, ; \bar{x}_2 = 4$
Then
$\sigma^2 = \dfrac{n_1\sigma_1^2 + n_2\sigma_2^2}{n_1+n_2} + \dfrac{n_1(\bar{x}_1 - \bar{x})^2 + n_2(\bar{x}_2 - \bar{x})^2}{n_1+n_2}$
where $\bar{x} = \dfrac{n_1\bar{x}_1 + n_2\bar{x}_2}{n_1+n_2} = 3$
$\Rightarrow \sigma^2 = \dfrac{5(4) + 5(5)}{10} + \dfrac{5(1)^2 + 5(1)^2}{10}$
$= 4.5 + 1 = 5.5 = \boxed{11/2}$
What is the probability of getting a sum 9 from two throws of dice?
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Solution
29/72
A random variable X follows a binomial distribution with parameters n = 5 and p = 0.3 . The expected value of X is:
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Solution
E(X)=5×0.3
E(X)=1.5
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