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## Homework Statement

Suppose an urn contains 6 white and 9 black numbered balls.

**a)**4 balls are collected in random without replacement. What is the probability the first two are white and the last two are black?

**b1)**4 balls are collected in random with replacement. What is the conditional probability the first and third balls are black, given that the quartet contains exactly 3 black balls.

**b2)**The same as b1, except now we collect the balls without replacement.

## Homework Equations

The conditional probability [tex] P(A|B) = \dfrac{P(A \cap B)}{P(B)} [/tex] and the permutations without replacement [tex] \dfrac{n!}{n1 \cdot n2 \cdot \cdot nk} [/tex]

## The Attempt at a Solution

**a)**I compute [tex] P(WWBB)=\dfrac{6}{15} \cdot \dfrac{5}{14} \cdot \dfrac{9}{13} \cdot \dfrac{8}{12} = 0.0659 [/tex]

**b1)**Suppose the events

A: the first and third balls are black

B: there are exactly 3 black balls in the quartet

The probability of B is:

[tex] P(B) = (\dfrac{9}{15})^3 [/tex] , since each time we take out a ball, we replace it, thus the probability is the same.

The probability of [tex] A \cap B [/tex] is:

[tex] P(A \cap B) = (\dfrac{9}{15})^3 \cdot (\dfrac{6}{15}) [/tex]

So, the conditional probability is:

[tex] P(A|B) = \dfrac{P(A \cap B)}{P(B)} = \dfrac{6}{15} [/tex]

**b2)**Here, the event B has two possibilities, either BBBW or BWBB. However, since we have multiplication, each case has the same probability, so:

[tex] P(B) = \dfrac{9}{15} \cdot \dfrac{8}{14} \cdot \dfrac{7}{13} \cdot \dfrac{6}{12} = 0.00923 [/tex]

The total number of ways that the 15 balls can be arranged is found by:

[tex] N=\dfrac{15!}{6! \cdot 9!} = 385 [/tex]

Since, the possible arrangements we want are BBBW and BWBB, I find:

[tex] P(A \cap B) = \dfrac{2}{385} = 0.0052 [/tex]

So, the conditional probability is:

[tex] P(A|B) = \dfrac{0.0052}{0.0923} = 0.0056 [/tex]

Is my solution correct?

Thank you.