Mht Cet

Probability

MHT CET 2024 9th May Evening Shift

MCQ+2 / -02024

A bag contains 4 red and 3 black balls. One ball is drawn and then replaced in the bag and the process is repeated. Let X denote the number of times black ball is drawn in 3 draws. Assuming that at each draw each ball is equally likely to be selected, then probability distribution of $X$ is given by

$x$	0	1	2	3
$\mathrm{P}(x)$	$\left(\frac{4}{7}\right)^3$	$\frac{9}{7} \cdot\left(\frac{4}{7}\right)^2$	$\frac{12}{7} \cdot\left(\frac{3}{7}\right)^2$	$\left(\frac{3}{7}\right)^3$

$x$	0	1	2	3
$\mathrm{P}(x)$	$\left(\frac{3}{7}\right)^3$	$\frac{12}{7} \cdot\left(\frac{3}{7}\right)^2$	$\frac{9}{7} \cdot\left(\frac{4}{7}\right)^2$	$\left(\frac{4}{7}\right)^3$

$x$	0	1	2	3
$\mathrm{P}(x)$	$\left(\frac{3}{7}\right)^3$	$\frac{9}{7} \cdot\left(\frac{4}{7}\right)^2$	$\frac{12}{7} \cdot\left(\frac{3}{7}\right)^2$	$\left(\frac{4}{7}\right)^3$

$x$	0	1	2	3
$\mathrm{P}(x)$	$\left(\frac{4}{7}\right)^3$	$\frac{12}{7} \cdot\left(\frac{4}{7}\right)^2$	$\frac{9}{7} \cdot\left(\frac{3}{7}\right)^2$	$\left(\frac{3}{7}\right)^3$

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