When estimating probability of 6 different outcomes, do I make my probability fraction from the frequency of each outcome, or by the possible outcomes. Example: A number cube was thrown 150 times. The results are shown in a table. Estimate the probability for each outcome. (table:) 1 = 33, 2 = 21, 3 = 15, 4 = 36, 5 = 27, and 6 = 18. So would the answer be 1/6 for each (probability) or would it be 1 = 33/150, 2 = 21/150, 3 = 15/150, ect.?

Question

anonymous · Answer

The outcome of a certain face appearing after a toss is independent of the results which came before it.  The prob. of any face appearing is (1/6), but the prob. of say the 1st face appearing twice in a row is (1/6)*(1/6)

anonymous · Answer

If the prob. was depended on what came before it (like drawing a King of Diamonds after a Queen of Spades was drawn), you would have to find the prob. differently.

anonymous · Answer

Well in this example there is no order of when each outcome was rolled

anonymous · Answer

The roll number doesn't matter, each face still has a (1/6) chance of being face up.  The number was just the total out of 150.  If you took all those numbers and found the expected value, it would be (1/6)

anonymous · Answer

If you want to show how many times out of the 150 each side was shown up, then you would use the 33/150, etc

anonymous · Answer

ok so in this problem which one would I need to show, 1/6 or 33/150?