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.
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:
4. 5 cards
Similarly, show that
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)=1-P(car)
Please check my calculations.
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
I am glad you checked things out.
For P(5), we have
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
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