Mht Cet

Probability

MHT CET 2021 23th September Morning Shift

MCQ+2 / -02021

The probability distribution of the number of doublets in four throws of
a pair of dice is given by

\(\mathrm{X}\)	0	1	2	3	4
\(\mathrm{P(X) :}\)	\(\frac{1}{5}\)	\(\frac{1}{5}\)	\(\frac{1}{5}\)	\(\frac{1}{5}\)	\(\frac{1}{5}\)

\(\mathrm{X}\)	0	1	2	3
\(\mathrm{P(X) :}\)	\(\frac{1}{4}\)	\(\frac{1}{4}\)	\(\frac{1}{4}\)	\(\frac{1}{4}\)

\(\mathrm{X}\)	1	2	3	4
\(\mathrm{P(X) :}\)	\(\frac{1}{4}\)	\(\frac{1}{4}\)	\(\frac{1}{4}\)	\(\frac{1}{4}\)

\(\mathrm{X}\)	0	1	2	3	4
\(\mathrm{P(X) :}\)	\(\frac{625}{1296}\)	\(\frac{125}{324}\)	\(\frac{25}{216}\)	\(\frac{5}{324}\)	\(\frac{1}{1296}\)

Continue this paper