question_answer

Find the probability distribution of number of doublets in three throws of a pair of dice.

Answer:

For a pair of dice, total number of event, \[n(S)=6\times 6=36\] Number of doublets = 6 Which are {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)} Let R = P (doublets in pair of dice) \[=\frac{6}{36}=\frac{1}{6}\] And Q = P (non-doublets in pair of dica) \[=1-\frac{1}{6}=\frac{5}{6}\] Let X denotes the number of doublets in pair of dica. Then, X can take value 0, 1, 2, 3. \[P(X=0)=P\](no doublet)             \[={{Q}^{3}}={{\left( \frac{5}{6} \right)}^{3}}=\frac{125}{216}\] \[P(X=1)=P\](one doublet and two non-doublets)             \[=QQR+RQQ+QRQ=3{{Q}^{2}}R\]             \[=3\times {{\left( \frac{5}{6} \right)}^{2}}\times \left( \frac{1}{6} \right)\]             \[=3\times \frac{25}{36}\times \frac{1}{6}=\frac{75}{216}\] \[P(X=2)=P\](two doublests and one non-doublet)             \[=RRQ+RQR+QRR\]             \[=3\,Q{{R}^{2}}=3\times \frac{5}{6}\times {{\left( \frac{1}{6} \right)}^{2}}\]             \[=3\times \frac{5}{6}\times \frac{1}{36}=\frac{15}{216}\] and \[P(X=3)=P\](three doublets)             \[={{R}^{3}}={{\left( \frac{1}{6} \right)}^{3}}=\frac{1}{216}\] \[\therefore \] required probability distribution is

X	0	1	2	3
P(X)	\[\frac{125}{216}\]	\[\frac{75}{216}\]	\[\frac{15}{216}\]	\[\frac{1}{216}\]