What is the probability

  • Thread starter bhoover05
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  • #26
tiny-tim
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Yes, I see how that works. I can NOT believe I was so easily confused by some basic algebra! AH! I feel quite ridiculous having stumbled through that the way I had. Thank you so much for your support!

hee hee! :biggrin:
For the question about the friends coming down to attend a football game, I do believe I have to use the equation. . .

There's no simple way of doing this …

the number of spare tickets wil be between 0 and 125 …

you want the probability that it's 4 or 5 or … or 125.

So it's best to calculate the probability that it's 0 or 1 or 2 or 3, and then subtract that from 1.

So, to start you off … what's the probability that it's 0, or 1? :smile:
 
  • #27
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Here I was, thinking I was on a roll. . . Ugh.

Well, would this involve using the same formula? The:

(n/x)p^x(1-p)^n-x

the p would still equal 0.05 since its 5%?
 
  • #28
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^ and the n would still be 125 since there are 125 students. . .
 
  • #29
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I have ran thru a couple of other problems that are a lot easier to see if I have improved at all. . . Thought maybe you would enjoy checking over them :-)?

1. A game of Yahtzee, you get 3 rolls of the dice per turn, and you get to choose which ones you re-roll each time. My brother calls me up and asks me the probability of him winning the game. If he gets a large straight, he wins, otherwise my Dad wins. With his first roll, he got a 2,3,4,6,6. He is going to re-roll the second 6 that he got, to try and get a 5. He has two rolls to get a 5. Whats the probability that he gets that 5 and beats my dad?
~~1st roll: fail to get a 5 w/ probability of (5/6)
probability to fail 2x's=(5/6)^2=0.694
Probability = 1-0.694= 0.3055556
Probability= 0.3056

2. You and I are playing a game. I'm going to be flipping a coin, and you will be rolling a die. I get to go first. If I get heads, I win the game, but if you roll the die and get the number 1,2,3,4, you win. It goes back and forth until someone wins. Who has the better chance of winning and why?
~~You automatically have a 0.5, or 50% chance of winning your coin toss on heads
My roll only takes place if you roll a tail
Your second flip only takes place if I lose my roll ( on a 5 or 6, thus a 1/3 probability)
Probability= E (1/2)^i (1/3)^i-1
(1/2)=coin
(1/3)=die
Probability for coin tossing player (you)= 0.6
Me= (rolling player) = 1-p= (1-0.6)= 0.4
My probability= 0.4 vs you with the 0.6 probability

3. (same set up as question above) BUT, I go first. If I get heads, I win. If not, you roll the die, and if you get the numbers 1,2,3,4,5 you win. If not, back to me, and if I get heads, I win. If not, you win by default. Who has a better chance of winning, and why?
~~P=E (1/2)^i (5/6)^i-1
(1/2)=coin
(5/6)=die
Assuming there is a 50% chance of winning going first, you still have a better chance of winning the game overall.

4. After a game, it was reported that of the 11 starting players, 8 hurt a hip, 6 hurt a arm, 5 hurt a knee, 3 hurt both a hip and an arm, 2 hurt both a hip and a knee, 1 hurt an arm and a knee, and no one hurt all three. Comment on the accuracy of the report.
~~For this one I drew a diagram and tried to assign the injuries to the players. When putting the knee injuries on the diagram, they would have to be overlaped to make the data work. This means that there would be injuries than listed, thus meaning the data is impossible.

5. Player A and B play a sequence of independent games. Player A throws a die first and wins on a 6. If he fails, player B throws and wins on a five or six. If he fails, A throws and wins on a four, five, or six. ( this continues in same pattern) Find the probability of each player winning.
~~A starts: win= (1/6)
lose=(5/6)
(1/6)+((5/6)*(4/6)*(3/6))+((5/6)*(4/6)*(3/6)*(2/6)*(1/6))
- Player A has a 52% probability
~~B goes second: win=(2/6)
lose=(4/6)
- Player B has a 48% chance

6. Five people are about to play Yahtzee. Whats the probability that I win the game?
~~ Since there are 5 people playing, there is a 1/5 chance of winning the game.
 

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