Posted by Mike jenkins on Wednesday, July 18, 2012 at 6:27pm.
cant tell if i am on right track for this prababilitly hw assigment help would good!
I was recently given a hw assigment dealing with the price is right's "money game". The contestant is shown nine cards on a board. Each card has one pair of numbers. One card has the first two digits of the price of the car and the second card has the last two digits of the price of the car. The middle digit is given. The contestant chooses one card at a time. If the card contains a pair of digits from a car price it is placed in one of the two card price slots. If it does not contain two pair of digits from the car price it is placed in one of the four money slots on the left of the board. The contestant wins the car if the two car slots are filled before the four money slots. Other wise the contestant eins a small amount of money. My teacher has asked me six questions 1) what is the probabililty of winning the car selecting two cards. 2) what is the probabililty winning the car selecting three cards. 3) what is the probabililty winning the car selecting four cards. 4) what is the probability winning the car selecting five cards. 5) what is the probabililty that the contestant will win a car 6) what is the probabilily the contestant will win money. I feel that when the teacher sais the term select it means i can do the ncr method because ncr method the numbers do not have to be in order and it seems like it is a without replacement type of questions. Im getting confused on what type of formula to use because i feel you can also put this in fraction form which give
You a decimal . I wanted to think Ican also use the permutation method but im not sure until this becomes clear to me. For the first question dealing with probabililty of winning car selecting two cards i wanted to use the fraction of 8/9 and 7/8 bu and
Muiltiply them and simplify but im not sure if my idea is right. There is the given probaililty idea that im thinking of like i sellect one tile and now it becomes 8/9 then second card makes it 7/8 i do mot know if im thinking wrong but if you can walk me theough this to gove me ways to think of this problem i havent thought of it really would help thanks

Statistics  MathMate, Wednesday, July 18, 2012 at 8:45pm
First I will paraphrase the game to make sure I understand the game the same way you're supposed to.
Contestant is to select randomly (assuming he doesn't see the digits while selecting) cards, one at a time, out of 9 cards (without replacement).
Two cards out of 9 are indicated to be related to a car (car cards). The remaining 7 simply called money cards.
If he picks the two car cards in 5 or less picks, he wins a car. This means he fills the two "car slots" before filling the four "money slots".
We define the following events:
C=selecting a car card
M=selecting a money card.
1. Contestant select two car cards in exactly two picks:
Winning event = CC
P(2)=P(CC)=(2 out of 9)(1 out of 8)=1/36
2. two car cards in exactly 3 picks
Winning events are CMC or MCC.
CCM is not considered because she would have won in the first two picks.
P(CMC)=(2/9)(7/8)(1/7)=1/36
P(MCC)=(7/9)(2/8)(1/7)=1/36
P(2)=P(CMC)+P(MCC)=1/18
3. Exactly 4 picks.
Since we have to leave the car card as the fourth pick, we have (3,1), or 3 choose 1 ways to arrange the first 3 cards, and as is is seen above, all three have the same probability, namely:
P(CMMC)=(2/9)(7/8)(6/7)(1/6)
P(MCMC)=(7/9)(1/8)(6/7)(1/6)
P(MMCC)=(7/9)(6/8)(2/7)(1/6)
so
P(3)=3(1)(2)(7)(6)/(9*8*7*6)
=3(1/36)=1/12
4. 5 cards
Similarly, show that
P(5)=4(1/36)=1/9
Finally,
Therefore the probability of getting a car equals
P(car) = P(2)+P(3)+P(4)+P(5)=10/36=5/18
(not bad for a game show).
The probability of getting money (i.e. not getting a car is P(money)=1P(car)
Please check my calculations.

Statistics  Mike jenkins, Thursday, July 19, 2012 at 9:27am
Thank you very much can you explain how you got the 10/36 i added all the p(2) through p(5) and got something different thanks again

Statistics  MathMate, Thursday, July 19, 2012 at 7:32pm
I am glad you checked things out.
For P(5), we have
P(CMMMC)=(2/9)(7/8)(6/7)(5/6)(1/5)=1/36
P(MCMMC)=(7/9)(2/8)(6/7)(5/6)(1/5)=1/36
The same goes for the rest, namely
MMCMC, MMMCC, which when added up together gives 4/36.
P(2) to P(3) have been calculated before.
For P(2) to P(5), I have
1/36+2/36+3/36+4/36
=10/36
=5/18
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