Probability Question

[DWG]

From EZ Series Advanced Workbook.

In a certain jar, there are 5 red marbles, 5 blue marbles, and 5 green marbles If two marbles are randomly drawn, what is the probability that the two marbles will be of different colors?

[Pravin]

I think here is the way to solve it -

1. First marble can be any color - so probability of selecting any one color marble will be - 1 (say it is red in color)

Now remaining red marbles are - 4 and total marbles are - 14.

So probability of not selecting red marble is - (14-4)/14 i.e. 10/14

resultant probability is - 1*10/14 = 5/7.

Hope this is correct. If there is a better faster way please suggest

[ahistegt4]

I think here is the way to solve it -

1. First marble can be any color - so probability of selecting any one color marble will be - 1 (say it is red in color)

Now remaining red marbles are - 4 and total marbles are - 14.

So probability of not selecting red marble is - (14-4)/14 i.e. 10/14

resultant probability is - 1*10/14 = 5/7.

Hope this is correct. If there is a better faster way please suggest

What you have done seems correct.

Total # of ways we can draw 2 marbles 15!/(2! 13!) = 105
# of ways two different colours are drawn 5*5+ 5*5+ 5*5=75

P(2 diff color)=75/105 = 5/7

[JonathanSchneider]

Well done both of you! While I like the second way a lot, remember that the first is much simpler for a problem such as this one.

[rahul.popat]

i didnt understand the following part of 2nd solution;
"# of ways two different colours are drawn 5*5+ 5*5+ 5*5=75 "

can som1 pls explain.

[twallach]

The 5 *5 thing works this way:

There are three color combos RB, RG, BG.

For each color, there are five marbles. For red: R1, R2, R3, R4, and R5

For each combo, there are thus 5 * 5, or 25, combos of marbles:

R1, B1
R1, B2
...
R5, B4
R5, B5

25 * 3 (because three combos) = 75.

Tommy Wallach
MGMAT Instructor

[atul.jadhav]

Hi All,

Is the following approach right?

Probability of selecting 2 marbles of different colors = 1 - total probability of selecting same color.

1.Probability of selecting both Red is (5/15)*(4/14) = 2/21
2. Total Probability of selecting Red,Green and Blue is (2/21)*3 = 2/7.

3. 1-2/7 = 5/7 is the probability of selecting 2 marbles with different colors

Thanks in advance!

Regards

Atul

[RonPurewal]

Hi All,

Is the following approach right?

Probability of selecting 2 marbles of different colors = 1 - total probability of selecting same color.

1.Probability of selecting both Red is (5/15)*(4/14) = 2/21
2. Total Probability of selecting Red,Green and Blue is (2/21)*3 = 2/7.

3. 1-2/7 = 5/7 is the probability of selecting 2 marbles with different colors

Thanks in advance!

Regards

Atul

that's good, too.

[mercytron]

SLOT METHOD ALTERNATIVE:
Possible outcomes using slot method:
Step 1:
15 ways of choosing marble 1 and 14 ways of choosing marble 2 i.e - 15 X 14.
Step 2:
Order does not matter - Divide by 2! to account for redundant pairs of marbles
Hence possible outcomes = (15 x 14)/2!

Desired outcomes:
Step 1:
15 ways of choosing marble 1 and 10 ways of choosing marble 2 of different color i.e - 15 X 10
Step 2:
Order does not matter - Divide by 2! to account for redundant pairs of marbles
Hence possible outcomes = (15 x 10)/2!

Answer: P(2 marbles of different colors)
= Desired outcomes/possible outcomes
= ((15 x 10)/2!)/((15 x 14)/2!)
=10/14
=5/7

[jnelson0612]

Looks good!