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.
To get the number of ways of getting two numbers the same and the other numbers different:
First use the pigeon hole principle: .
Now arrange the above: .
So the number of ways of getting two numbers the same and the other numbers different is equal to 240.
Okay, let's baby-step through it . . .
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: . choices.
The first die can have any number: .Prob =
The second die must have the same number: .Prob =
The third die must have a different number: .Prob =
The fourth die must have yet another different number: .Prob =
Therefore, the answer is: .
First of all, , 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".