Respuesta :
Answer:
Following are the answer to this question:
Step-by-step explanation:
In the question first calls the W if the transmitted chip was white so, the W' transmitted the chip is red or R if the red chip is picked by the urn II.
whenever a red chip is chosen from urn II, then the probability to transmitters the chip in white is:
[tex]P(\frac{w}{R}) = \frac{P(W\cap R)}{P(R)} \ \ \ \ \ _{Where}\\\\P(R) = P(W\cap R) + P(W'\cap R) \\[/tex]
The probability that only the transmitted chip is white is therefore [tex]P(W) = \frac{2}{3}\\[/tex], since urn, I comprise 3 chips and 2 chips are white.
But if the chip is white so, it is possible that urn II has 4 chips and 2 of them will be red since urn II and 2 are now visible, and it is possible to be: [tex]P(\frac{R}{W}) = \frac{2}{3}[/tex]
[tex]P(W\cap R) = P(W) \times P(\frac{R}{W}) \\[/tex]
[tex]= \frac{2}{3}\times \frac{2}{4} \\\\= \frac{2}{3}\times \frac{1}{2} \\\\= \frac{2}{3}\times \frac{1}{1} \\\\=\frac{1}{3}\\\\= 0.333[/tex]
Likewise, the chip transmitted is presumably red [tex](P(W')= \frac{1}{3})[/tex]and the chip transferred is a red chip of urn II [tex](P(\frac{R}{W'})= \frac{3}{4}[/tex], and a red chip is likely to be red [tex](\frac{R}{W'})[/tex].
Finally, [tex]P(W'\cap R) = P(W') \times P(\frac{R}{W'})\\[/tex]
[tex]= \frac{1}{3} \times \frac{3}{4} \\\\ = \frac{1}{1} \times \frac{1}{4} \\\\=\frac{1}{4}\\\\= 0.25[/tex]
The estimation of [tex]P(R)[/tex] and [tex]P(\frac{W }{R})[/tex] as:
[tex]P(R) = 0.3333 + 0.25\\\\ \ \ \ \ \ \ \ \ \ = 0.5833 \\\\ P(\frac{W}{R}) = \frac{0.3333}{0.5833} \\\\\ \ \ \ \ \ \ = 0.5714[/tex]
