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A biased coin, twice as likely to come up heads as tails, is tossed once. If it shows heads, a chip is drawn from urn I, which contains three white chips and four red chips; if it shows tails, a chip is drawn from urn II, which contains six white chips and three red chips. Given that a white chip was drawn, what is the probability that the coin came up tails.

Respuesta :

Manetho

Answer:

[tex]=\frac{7}{16}}[/tex]

Step-by-step explanation:

[tex]\mathrm{Let\;me\;denote\;Probability\;by\;P\;through\;out\;my\;answer}[/tex]

[tex]\mathrm{Heads\rightarrow P(H)=\frac{2}{3}}[/tex]

[tex]\mathrm{Head\Rightarrow Urn\;1}[/tex]

[tex]\mathrm{P(White\;chip| Heads)=\frac{3}{3+4}=\frac{3}{7}}[/tex]

[tex]\mathrm{Tails\rightarrow P(T)=\frac{1}{3}}[/tex]

[tex]\mathrm{P(White\;chip| Tails)=\frac{6}{3+6}=\frac{6}{9}}[/tex]

[tex]\mathrm{We\;now\;want\;P(Tails|White\;Chip)}[/tex]

[tex]\mathrm{\;P(Tails|White\;Chip)=\frac{P(White\;Chip|Tails)P(Tails)}{P(White\;Chip|Tails)P(Tails)+P(White\;Chip|Heads)P(Head}[/tex]

[tex]\mathrm{\Rightarrow \;P(Tails|White\;Chip)=\frac{\left ( \frac{6}{9} \right )\left ( \frac{1}{3} \right )}{\left ( \frac{6}{9} \right )\left ( \frac{1}{3} \right )+\left ( \frac{3}{7} \right )\left ( \frac{2}{3} \right )}}[/tex]

[tex]\mathrm{\Rightarrow \;P(Tails|White\;Chip)=\frac{\frac{2}{3}}{\frac{2}{3}+\frac{6}{7}}=\frac{14}{32}=\frac{7}{16}}[/tex]

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