# Probability Question

• Dec 16th 2008, 08:36 AM
feage7
Probability Question
5 balls are picked at random one at a time and then replaced. If 65% of balls are blue, 30% are red and 5% are green. Then what is the probability that
i) All 5 balls are red
ii) 3 are blue, 1 is green and 1 is red
iii) At least 1 is a green
• Dec 16th 2008, 08:40 AM
colby2152
Quote:

Originally Posted by feage7
5 balls are picked at random one at a time and then replaced. If 65% of balls are blue, 30% are red and 5% are green. Then what is the probability that
i) All 5 balls are red
ii) 3 are blue, 1 is green and 1 is red
iii) At least 1 is a green

BTW, if the balls were not replaced, then this problem would be unsolvable due to the unknown quantity of balls.

i) $\displaystyle P(red)^5 = 0.3^5$
ii) $\displaystyle P(blue)^3 * P(green) * P(red) = 0.65^3 * 0.3*0.05$
iii) $\displaystyle P(G \ge 1) = 1 - P(ge = 0) = 1 - 0.95^5$
• Dec 16th 2008, 05:14 PM
mr fantastic
Quote:

Originally Posted by colby2152
BTW, if the balls were not replaced, then this problem would be unsolvable due to the unknown quantity of balls.

i) $\displaystyle P(red)^5 = 0.3^5$
ii) $\displaystyle P(blue)^3 * P(green) * P(red) = 0.65^3 * 0.3*0.05$ Mr F says: I think this answer might need to be multiplied by $\displaystyle {\color{red}\frac{5!}{3! \, 1! \, 1!}}$.
iii) $\displaystyle P(G \ge 1) = 1 - P(ge = 0) = 1 - 0.95^5$

To the OP: Read Multinomial distribution - Wikipedia, the free encyclopedia
• Dec 19th 2008, 09:18 PM
xzed
Mr. F, I don't understand why you multiplied it by 20?
because if you pick, that's a combination, so the arrangement shouldn't matter, I might be wrong, since I am new to probability
• Dec 20th 2008, 02:15 AM
Pim
The reason it has to be multiplied is exactly because order does not matter.
$\displaystyle P(blue)^3 * P(green) * P(red) = 0.65^3 * 0.3*0.05$

Assumes that you first pick three blue balls, then a green one, then a red one. Because order does not matter, you'll need to multiply this by the amount of combinations possible, which in this case is:
$\displaystyle {\color{red}\frac{5!}{3! \, 1! \, 1!}}$