Email About #4...

Here's a response I had to a question about number 4...hopefully it helps with your work!


Although I played more than I should for the first half of the week, I have taken a significant break from the action to get all of your papers graded.

For your work, you can just use your individual numbers for number four. However, on #4 you need not use your mean. On 4a, to find the probability of no winning numbers simply count up how many "0's" you have and divide by the total number of rolls in your simulation (that's where the x/40 is coming from). Similarly, on 4b, count how many "1's" you had in your table and divide this number by 40. Finally, for 4c, count how many "2's" you have and divide by 40. Again, use the same logic for 4d, counting how many "3s," which is like 0 total (maybe 1 or 2).

Now, for 4, it calls for two answers--the theoretical probability and using your data. The process above is the process to find probability for your data. For the theoretical, consider the following:

4a: What is the probability of "no winning numbers?" You rolled 3 dice in this game, so that would mean (NOT WINNING and NOT WINNING and NOT WINNING). What is the probability of "NOT WINNING?" In other words, how many total numbers were NOT the winning number you chose? Out of how many total numbers on a die? And then, think--what does "and" mean in terms of mathematical operations?

4b: Now we have the probability of "1 winning number." This means WINNING and NOT WINNING and NOT WINNING. We have to do a little more though--is this the only way you can have exactly one winning number show up? Does it have to be first? We'll have to multiply this first answer by the number of combinations (the number of ways we can arrange exactly 1 WINNING).

4c: Similar to b. Remember for to account for all the ways to arrange two WINNING now.

4d: You got this.

And that is a crazy answer for 4a. Try again.

See you Monday! :)