Multinomial Distribution

Probability Level 3

An urn contains 1 blue, 1 green and 1 yellow ball. With replacement, we draw out a ball 7 times and record down the color seen. What is the probability that we see an equal number of blue and green balls?

The answer can be written as $\dfrac{a}{b}$ with $a$ and $b$ coprime positive integers. Submit $a+b$ .

The answer is 860.

2 solutions

Guillermo Templado
Mar 11, 2016

The probability is: $(\frac{1}{3})^7 \cdot (\frac{7!}{0!0!7!} + \frac{7!}{1!1!5!} + \frac{7!}{2!2!3!} + \frac{7!}{3!3!1!})=$ $= (\frac{1}{3})^7 \cdot (1 + 42 + 210 + 140) = (\frac{1}{3})^7 \cdot (393) = \frac{131}{729} = \frac{D}{E} \Rightarrow D + E = 860$

Great question, the scenarios where the blue equals green can only happen if the yellow ball shows up 7 times(7 choose 7), 5 times (7 choose 5), 3 times (7 choose 3), and 1 time (7 choose 1) respectively. Then you have to multiply the scenarios each by combinations of placing the blue and green balls into the rest of them. For the 5 times Y ball, combination of 1 G and 1 B is (2 choose 1), 2 G and 2 B is (4 choose 2) and lastly you have 3G and 3B (6 choose 3).

Harry Song - 5 years, 2 months ago

Great solution using Permutations with Repetition.

Tran Quoc Dat - 5 years, 1 month ago

Alex Li
Mar 18, 2016

P(x) = x green and x blue balls * 3^7 (total number of possibilities).

P(0) = 1 (duh)

P(1) = 5 yellow balls, so 6 choose 1 places the first color could go ( 5 starting balls + 1 ball being "placed" in the set), 7 choose 1 places the second could go from the next set.

P(2) = 3 yellow balls, so 5 choose 2 places the first color could go, 7 choose 2 places the second could go from the next set.

P(3)= 1 yellow ball, so 4 choose 3 places the first color could go, 7 choose 3 places the second could go from the next set.

1 + 6C1 * 7C1+5C2 * 7C2+4C3 * 7C3 = 393

393/2187 = 131/729

131+729= 860

Multinomial Distribution

The answer is 860.

2 solutions

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