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Bowl I contains 7 red and white chips and bowl II has 4 red and 6 white chips. Two chips are selected at random and without replacement from I and transferred to II. Three chips are then selected at random and without replacement from II.

a. What is the probability that all three are white? b. Given that three white chips are selected from II, what is the conditional probability that two white chips were transferred from I?

Jonarie Ramos Vergara
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2 Answers2

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Hint for (a): use the law of total probability, conditioning over which two chips are transferred.

Then for (b), Bayes' theorem gives \begin{equation} P(\text{2 transferred} \mid \text{3 selected}) = \frac{P(\text{3 selected} \mid \text{2 transferred})P(\text{2 transferred})}{P(\text{3 selected})}. \end{equation} The two terms in the numerator should be easy; the denominator is your answer from part (a).

Platehead
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a) What are all of the possibilities for transferring to bowl II?

  1. Both chips are white (what is the probability)?
  2. Both chips are red (what is the probability)?
  3. One chip is red and one is white (what is the probability)?

From each of those possibilities you will end up (in bowl II) with either

  1. 8 white and 4 red (if you transfer two whites form bowl I to bowl II)
  2. 6 white and 6 red (if you transfer two red from bowl I to bowl II)
  3. 7 white and 5 red (if you transfer 1 white and 1 red from bowl I to bowl II).

Calculate the probabilities of selecting three white from each of those cases then multiple each of those cases by the probability of each case:

  • p(draw 3 white from II given transfer two whites from I to II)
  • p(draw 3 white from II given transfer two red from I to II)
  • p(draw 3 white from II given transfer one white and one red from I to II))

Then add up the probability of each case (you can add because you cannot both pick two white chips and pick one white chip and one red, etc.).

b) You already have the probability of three white chips being selected (from part a)). Using Baye's theorem, you can compute the conditional probability:

$$ P(A|B) = \frac{P(B|A)\cdot P(A)}{P(B)} $$

Here 'A' is "two white chips were transferred from I to II" (we know this probability: $P(A) = \frac{7\cdot6}{14\cdot 13} = \frac{\binom{7}{2}}{\binom{14}{2}}$. $B$ in this case is that "three white chips are selected from II". We certainly know $P(B)$ from part a) (we already calculated that). In addition we only need $P(B | A)$ which should be part of your calculation for part a). That is, if we assume two white chips were transferred to II, then we have 8 white chips and 4 red chips thus the probability of choosing two white chips is $P(B|A) = \frac{8*7}{12*11} = \frac{\binom{8}{2}}{\binom{12}{2}}$

Jared
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  • p(draw 3 white from II given transfer two whites from I to II)$ = = \frac {\frac{\binom{8}{3}\binom{4}{0}}{\binom{12}{3}}\frac{\binom{3}{2}\binom{7}{0}}{\binom{10}{2}}} $ – Jonarie Ramos Vergara Oct 04 '14 at 07:57
  • You should fix your latex, but you shouldn't have that many combinations anyway. If we assume two white go to II, then we have (in II) 4 red and 8 whites. The probability of drawing three white is then: $\frac{876}{121110} = \frac{\binom{8}{3}}{\binom{12}{3}}$. – Jared Oct 04 '14 at 08:08