Probability

5/12

Consider the possibile outcome of two rolls of a die (where order matters):

(1,1)

(1,2)

(1,3)

.

.

.

(4,6)

(5,6)

(6,6)

There are 6*6=36 possible outcomes. Of these, the following would achieve the result desired:

(1,2)

(1,3)

(1,4)

(1,5)

(1,6)

(2,3)

(2,4)

(2,5)

(2,6)

.

.

.

(5,6)

For the first die equal to 1, there are 5 successful results. For the first die equal to 2, there are 4 successful results....For the first die equal to 5, there is 1 successful result. The total number of successful outcomes=5+4+3+2+1.

So the probability that the desired outcome is obtained=(5+4+3+2+1)/36=15/36=5/12.

The lowest number I can roll on a die is 1.

In ten rolls, that's a minimum of ten; I cannot make it 8 or less.

Zero is a foul throw.

If the coin is fair and balanced, like Fox, then the probability is 50%.

1/2*25/51*24/50*23/49

if cards are not replaced

4662

s

Most people take samples so that they may make estimates of parameters of interest: mean, variance, etc for the whole population. For such an estimate to have any validity the sample data must be assumed to represent a population distribution. Otherwise any conclusions based on the sample are valid only for the sample: hardly worth the effort!

yes.

one out of four

Sigurd Roll died on 1944-06-24.

In the first two draws, the probability is 1/15.

Sana could have 80.

They don't see the whole picture.

One side of the six sided cube has a 3 so: 1/6

One in 52 - because there are 52 cars in a deck, and only one Queen of Clubs.

unsay result for tomorrow jai alai january 7 2012

7

Yes.

1/5

1 out of 5.

720

No it can not; probability must be between 0 and 1, inclusive.

It's doesn't appear to be, because the probabilities don't add to 1.

In a straight line, 21! = 51,090,942,171,709,400,000 or approximately 51 quintillion.

Yes