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Home  >>  CBSE XII  >>  Math  >>  Probability
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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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  • 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
(\(X\;0\;1\;2\))
(\(p(x)\;0.54\;0.42\;0.04\))

 

answered Feb 28, 2013 by poojasapani_1
edited Jun 4, 2013 by poojasapani_1
 

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