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# Two biased dice are thrown together.For the first die $P(6)=\Large \frac{1}{2}$,the other scores being equally likely while for the second die,$P(1)=\Large \frac{2}{5}$ and the other scores are equally likely.Find the probability distribution of 'the number of ones seen'.

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A)
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• Two biased setting dice
• 1$$^{st}$$ dice P(6)=$$\frac{1}{2}$$
• others are equally likely
• P(1)=$$p(2)$$=$$p(3)$$=$$p(4)$$=$$p(5)$$=$$\large\frac{1}{2}$$/5=$$\large\frac{1}{10}$$
• 2$$^{nd}$$ dice$$P(1)$$=$$\large\frac{2}{5}$$ others eqally likely
• $$p(other\; than \;one)$$=$$\large\frac{3}{5}$$
• X denots number of 'Y' seen
• X=($$0\;1\;2)$$ when the two die are thrown togather
=$$P(X=0)$$=P(neither shows 1)
=P(1$$^{st}$$shows$$6\;5\;4\;3\;2\; and\; 2^{nd} shows\; other\; than\; one$$)
=($$\large\frac{1}{2}$$+$$\large\frac{4}{10}$$)$$\times\large\frac{3}{5}$$=$$\large\frac{27}{50}$$=$$0.54$$
=$$P(X=1)$$=P(either shows 1)
=P(1$$^{st}$$shows1 and 2$$^{nd}$$shows other)+P(1$$^{st}$$shows other 2 shows'1')
=($$\large\frac{1}{10}\;\times\;\frac{3}{5}\;+\frac{4}{10}\;+\;\frac{1}{2}\;\times\;\frac{2}{5}$$)
=$$\large\frac{3}{50}\;+\;\frac{6}{10}\;=\;\frac{33}{50}\;=(0.42) \(p(x=2)$$=P(both shows '1')
=$$\large\frac{1}{10}$$$$\times\large\frac{2}{5}$$=$$\large\frac{2}{50}$$
=$$0.04$$
Probability distrubation is
($$X\;0\;1\;2$$)
($$p(x)\;0.54\;0.42\;0.04$$)