# probability using permutations/combinations

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• July 10th 2007, 06:03 PM
sam48
probability using permutations/combinations
A bag contains 4 red and 3 green blocks. 3 blocks are randomly selected from the bag. Determine the probability that all three are red given that

i.) replacement occurs between the second and third selections

ii.) no replacement occurs
• July 10th 2007, 06:59 PM
Soroban
Hello, Sam!

Quote:

A bag contains 4 red and 3 green blocks.
Three blocks are randomly selected from the bag.
Determine the probability that all three are red given that:

a) replacement occurs between the second and third selections

First block is red: $P(\text{\#1 red}) = \frac{4}{7}$
Second block is red: $P(\text{\#2 red}) = \frac{3}{6}$ . . . . The second block replaced.
Third block is red: $P(\text{\#3 red}) = \frac{3}{6}$

Therefore: $P(\text{3 red}) \;=\;\frac{4}{7}\cdot\frac{3}{6}\cdot\frac{3}{6} \;=\;\frac{1}{7}$

Quote:

b) no replacement occurs
First block is red: $P(\text{\#1 red}) = \frac{4}{7}$
Second block is red: $P(\text{\#2 red}) = \frac{3}{6}$
Third block is red: $P(\text{\#3 red}) = \frac{2}{5}$

Therefore: $P(\text{3 red}) \;=\;\frac{4}{7}\cdot\frac{3}{6}\cdot\frac{2}{5} \;=\;\frac{4}{35}$
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