MAH CET MCA Probability Distribution Previous Year Questions (PYQs)
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Showing 2 questions
The probability distribution function of a random variable $X$ is given by
$f(x) = \dfrac{x}{18}, ; 0 \le x \le 6$
$= 0, \text{ otherwise}$
Then the value of $P(X > 2)$ is
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Solution
$P(X > 2) = \int_{2}^{6} \dfrac{x}{18} , dx = \dfrac{1}{18} \left[\dfrac{x^2}{2}\right]_2^6 = \dfrac{1}{36}(36 - 4) = \dfrac{32}{36} = \dfrac{8}{9}$
If, for a binomial distribution, the number of trials is $9$, the variance is $2$ and the probability of success is greater than that of failure, find the probability of both success.
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Solution
For a binomial distribution, variance $= npq = 2$ and $n = 9$.
So, $9p(1 - p) = 2 \Rightarrow 9p - 9p^2 = 2 \Rightarrow 9p^2 - 9p + 2 = 0$.
$\Rightarrow p = \dfrac{9 \pm \sqrt{81 - 72}}{18} = \dfrac{9 \pm 3}{18}$.
Hence, $p = \dfrac{2}{3}$ or $\dfrac{1}{3}$.
Since probability of success is greater than failure, $p = \dfrac{2}{3}$.
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