Algebra 2 probability question

Apr 2010
5
0
there are a bag of 7 marbles, 3 red 4 green

if u pick 2 marbles, what is the probability of choosing one green and one red (order doesnt matter)?

isn't it 3/7 x 4/6?

i kno 4 nCr 1 x 3 nCr 1 divided by 7 nCr 2 is the correct answer but how is it different from 3/7 x 4/6?
 

Soroban

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May 2006
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Hello, mairjuanaman!

There are a bag of 7 marbles: 3 red, 4 green.

if u pick 2 marbles, what is the probability of choosing one green and one red (order doesn't matter)?

Isn't it 3/7 x 4/6 ? . . . . . no

You found the probability of getting a Red, then a Green . . . in that order.

\(\displaystyle P(\text{Red, then Green}) \:=\:\frac{3}{7}\times\frac{4}{6} \:=\:\frac{2}{7}\)

And: .\(\displaystyle P(\text{Green, then Red}) \:=\:\frac{4}{7}\times\frac{3}{6} \:=\:\frac{2}{7}\)


Therefore: .\(\displaystyle P(\text{Red and Green, either order}) \;=\;\frac{2}{7} + \frac{2}{7} \;=\;\frac{4}{7}\)


This is the same answer as: .\(\displaystyle \frac{(_4C_1)(_3C_1)}{_7C_2} \)