• Apr 23rd 2009, 08:39 PM
Jason2009
Four dice are tossed, what is the probability of getting one pair and 2 other single number?

I try 6*5C2*????
That is all i got so far and I dont know how to figure out the rest.

(Crying)
• Apr 24th 2009, 01:39 AM
mr fantastic
Quote:

Originally Posted by Jason2009
Four dice are tossed, what is the probability of getting one pair and 2 other single number?

I try 6*5C2*????
That is all i got so far and I dont know how to figure out the rest.

(Crying)

You need the number of ways of getting two numbers the same and the other numbers different divided by the total number of ways.

To get the number of ways of getting two numbers the same and the other numbers different:

First use the pigeon hole principle: $\displaystyle 1 \times 1 \times 5 \times 4 = 20$.

Now arrange the above: $\displaystyle \frac{4!}{2!} = 12$.

So the number of ways of getting two numbers the same and the other numbers different is equal to 240.
• Apr 24th 2009, 01:50 AM
Jason2009
Really confused.
• Apr 24th 2009, 01:55 AM
mr fantastic
Quote:

Originally Posted by Jason2009
Really confused.

Please post something that makes sense. What do you mean by "No"? Where has 720 come from? State clearly what you don't understand.
• Apr 25th 2009, 02:19 PM
Jason2009
The answer of this question is 720 , and I have no idea how to figure it out.
240 is not right.
• Apr 25th 2009, 02:56 PM
Soroban
Hello, Jason2009!

Okay, let's baby-step through it . . .

Quote:

Four dice are tossed, what is the probability of getting one pair and 2 single numbers?

Two of the dice will have the same number . . . which two?
. . There are: .$\displaystyle {4\choose2} \:=\:{\color{blue}6}$ choices.

The first die can have any number: .Prob = $\displaystyle {\color{blue}\frac{6}{6}}$

The second die must have the same number: .Prob = $\displaystyle {\color{blue}\frac{1}{6}}$

The third die must have a different number: .Prob = $\displaystyle {\color{blue}\frac{5}{6}}$

The fourth die must have yet another different number: .Prob = $\displaystyle {\color{blue}\frac{4}{6}}$

Therefore, the answer is: .$\displaystyle 6\cdot\frac{6}{6}\cdot\frac{1}{6}\cdot\frac{5}{6}\ cdot\frac{4}{6} \;=\;\frac{720}{1296} \;=\;\boxed{{\color{blue}\frac{5}{9}}}$

• Apr 25th 2009, 04:08 PM
mr fantastic
Quote:

Originally Posted by mr fantastic
You need the number of ways of getting two numbers the same and the other numbers different divided by the total number of ways.

To get the number of ways of getting two numbers the same and the other numbers different:

First use the pigeon hole principle: $\displaystyle 1 \times 1 \times 5 \times 4 = 20$.

Now arrange the above: $\displaystyle \frac{4!}{2!} = 12$.

So the number of ways of getting two numbers the same and the other numbers different is equal to 240.

Well, I got careless here. Nevertheless, with a small bit of thought it should have been clear how to fix things up.

First of all, $\displaystyle \frac{4!}{2!} = 6$, not 12. You should have spotted that carelessness straight off. Then you multiply by 6 for the 6 possible numbers. So you end up with ..... 720, not 240.

See, all it takes is a bit of willingness to engage with and try to understand what gets posted, instead of saying "240 isn't right and I don't know what to do".