CA super lotto problem
Problem Statement: 
The California super lotto problem has asked us to figure out how many different combination of numbers there are, what’s the probability of winning and what are your expected winnings if the payout is $8,000,000 if you math all 5 numbers and the mega number with the cost of $1.

Process statement:On the first day of second semester we were given this sheet titled "California super lotto problem" and 2 points extra credit. We were told to write down 5 numbers between 147 (the numbers cannot repeat) then a mega number between 127 (may be a repeat from one of the other 5 numbers). If our numbers matched the random number generator then we get $20, but to get a chance to get that twenty we had to give up our extra credit points. I didn't need the extra 2 points but I decided not to play anything but 'what's the possibility of me getting the exact number?' I thought. On question four on the page we had to guess the odds of winning 1 in blank. I put 1 in the total amount of people who play just because I didn't know what number to put down.
How many different number combinations are possible for a CA super lotto ticket?So there are 5 numbers from 147 and mega number from 127. The 5 numbers cannot repeat itself but the mega number may be a repeat from one of the others. I multiplied 47 by 46 by 45 by 44 by 43 by 27 because the sample space will go down by one, by the fifth number picking you would have already picked 4 other numbers so 474=43. And 27 is stays as 27 because the number can be a repeat from one of the others that's already been chosen.
47*46*45*44*43*27=4,969,963,360 which is the total combinations. 

What is the probability of winning the CA Super lotto?
To win the California Super Lotto you have to have all 5 numbers (does NOT have the be in the exact order) and the mega number. We already know the sample spaces 47*46*45*44*43*27=4,969,962,360. If we have to make the five number in any order the 47 will be 5/47, any of the 5 numbers has to match the first number. The second number will be 4/46 because we already have the first number so the sample space already went down and now the four remaining numbers has to make the second number. Then it will be 3/45 then 2/44 then 1/43. The mega number is 1/27 because you have to get the exact number, if not you didn't win the jackpot.
Now we have 5/47*4/46*3/45*2/44*1/43*1/27=120/4,969,962,360 that simplifies to 1/41,416,353.
Now we have 5/47*4/46*3/45*2/44*1/43*1/27=120/4,969,962,360 that simplifies to 1/41,416,353.
If you match all 6 numbers, you win $8,000,000. It costs $1 to play, what are your expected winnings?
I used the expected value formula. I plugged in the money you would win so far I have E(x)=8,000,000 then we put in parenthesis the probability of winning which is 1/41,416,353. So now we have E(x)=8,000,000(1/41,416,353) now we add the probability of losing so now we have E(x)=8,000,000(1/41,416,353)+(1)(41,416,352/41,416,353).
Then we get 8,000,000/41,416,35341,416,352/41,416,353=33,416,352/41,416,353=.806839559. So basically you lose .81 cents each ticket you buy.
Then we get 8,000,000/41,416,35341,416,352/41,416,353=33,416,352/41,416,353=.806839559. So basically you lose .81 cents each ticket you buy.
Problem evaluation: 
Self evaluation: 
Edits: 
Yes I did liked this problem a lot. It did push my thinking and I felt like I did learn a lot of in this unit. I never really learned probability expect some really basic things so this really helped me understand.

On this problem I felt like I grew. In the past, I have been a really individual worker and I never really tried to include my table mates. In this problem I struggled at some points trying to understand it, so I asked my table for help. I learned that asking for help is okay.

Edits from me peers:
 fix photos  separate numbers more  add to process statement 