Question

urn I contains 1 red chip and 2 white chips; urn II contains 2 red chipsand 1 white chip. One chip is drawn at random from urn I and transferred to urnII. Then one chip is drawn from urn II. Suppose that a red chip is selected from urnII. What is the probability that the chip transferred was white

Answer 1

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:  

$P(\frac{w}{R}) = \frac{P(W\cap R)}{P(R)} \ \ \ \ \ _{Where}\\\\P(R) = P(W\cap R) + P(W'\cap R)$

The probability that only the transmitted chip is white is therefore $P(W) = \frac{2}{3}$, 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: $P(\frac{R}{W}) = \frac{2}{3}$

$P(W\cap R) = P(W) \times P(\frac{R}{W})$

                $= \frac{2}{3}\times \frac{2}{4} \\\\= \frac{2}{3}\times \frac{1}{2} \\\\= \frac{2}{3}\times \frac{1}{1} \\\\=\frac{1}{3}\\\\= 0.333$

Likewise, the chip transmitted is presumably red $(P(W')= \frac{1}{3})$and the chip transferred is a red chip of urn II $(P(\frac{R}{W'})= \frac{3}{4}$, and a red chip is likely to be red $(\frac{R}{W'})$.  

Finally, $P(W'\cap R) = P(W') \times P(\frac{R}{W'})$

                              $= \frac{1}{3} \times \frac{3}{4} \\\\ = \frac{1}{1} \times \frac{1}{4} \\\\=\frac{1}{4}\\\\= 0.25$

The estimation of $P(R)$ and $P(\frac{W }{R})$ as:  

$P(R) = 0.3333 + 0.25\\\\ \ \ \ \ \ \ \ \ \ = 0.5833 \\\\ P(\frac{W}{R}) = \frac{0.3333}{0.5833} \\\\\ \ \ \ \ \ \ = 0.5714$