# Math Help - Perms and Combs... a nightmare!

1. ## Perms and Combs... a nightmare!

Hey

Probability is such a nightmare for me, here's a two part Q I'm struggling with.

A jar contains 5 red, 2 blue and 3 green marbles. What is the probablity of choosing:
a) at least one green and at least one red marble
b) at least one red given that at least one green is chosen

For part B I'm using the formula of Pr(A|B) = Pr(A int B) / (Pr B), but I cannot for the life of me figure out how to get the intersection of "at least one red" and "at least one green".

Help would be much appreciated, thanks all!

2. ## Re: Perms and Combs... a nightmare!

Originally Posted by Oaky
Hey

Probability is such a nightmare for me, here's a two part Q I'm struggling with.

A jar contains 5 red, 2 blue and 3 green marbles. What is the probablity of choosing:
a) at least one green and at least one red marble
b) at least one red given that at least one green is chosen

For part B I'm using the formula of Pr(A|B) = Pr(A int B) / (Pr B), but I cannot for the life of me figure out how to get the intersection of "at least one red" and "at least one green".

Help would be much appreciated, thanks all!
For a) you don't need to worry about intersections. Just use a tree diagram.

In fact, you can use the tree diagram for part b) as well. Just disregard everything that doesn't start with a green...

3. ## Re: Perms and Combs... a nightmare!

Originally Posted by Oaky
Hey

Probability is such a nightmare for me, here's a two part Q I'm struggling with.

A jar contains 5 red, 2 blue and 3 green marbles. What is the probablity of choosing:

a) at least one green and at least one red marble

Help would be much appreciated, thanks all!

(10 = the possible outcomes, 9 = possible outcome after first event, providing they're not independent events)

3 x 5
10 9

= 15
90

= 1
6

4. ## Re: Perms and Combs... a nightmare!

i believe this is only half the story.

yes, the chances of selecting a green marble, then a red marble is 1/6.

but the chance of selecting a red marble, then a green marble is (5/10)(3/9) = 1/6,

so the chance of either happening (as these are mutually exclusive chains of events) is 1/6 + 1/6 = 1/3.

also, there are 10!/(2!8!) = 45 ways to pick 2 marbles out of 10.

there are (5)(3) = 15 ways this pair could be a red and green pair. therefore, we have 15/45 = 1/3 chance of getting a red and green pair.

5. ## Re: Perms and Combs... a nightmare!

Hello, Oaky!

Didn't anyone notice that the problem is incomplete?

A jar contains 5 red, 2 blue and 3 green marbles.
What is the probablity of choosing:

(a) at least one green and at least one red marble.

(b) at least one red, given that at least one green is chosen.
How many marbles are chosen?

Suppose 3 marbles are chosen.

Then there are: . ${10\choose3} \,=\,120$ possible outcomes.

(a) At least one G and at least one R.

GRR: . ${3\choose1}{5\choose2} \,=\,30$
GRG: . ${3\choose2}{5\choose1} \,=\,15$
GRB: . ${3\choose1}{5\choose1}{2\choose1} \,=\,30$

Hence, there are: . $30+15+30 \,=\,75$ ways with at least 1 G and at least 1 R.

$P(G \ge 1\:\wedge\:R \ge 1) \:=\:\frac{75}{120} \:=\:\frac{5}{8}$

(b) At least 1 Red, given at least 1 G.

We have: . $P(R \ge1\,|\,G \ge 1) \;=\;\frac{P(R \ge 1\:\wedge\:G \ge 1)}{P(G \ge 1)}$

We found the numerator in part (a): . $P(R \ge 1\:\wedge\:G \ge 1) \:=\:\frac{5}{8}$

The denominator is:. $P(\text{at least 1 G})$

. . The opposite is:. $P(\text{no G})$

. . There are: . ${7\choose3} \,=\,35$ ways to get no G.

. . Then:. $P(\text{no G}) \:=\:\tfrac{35}{120} \:=\:\tfrac{7}{24}$

. . Hence: $P(\text{at least 1 G}) \:=\:1-\tfrac{7}{24} \:=\:\tfrac{17}{24}$

Therefore: . $P(R\ge1\,|\,G\ge1) \;=\;\dfrac{\frac{5}{7}}{\frac{17}{24}} \;=\;\frac{15}{17}$