$Probability/Outcome$

Question

$Probability/Outcome$

A complete revision:

Match and Win Promotion

A pre-selected envelope will contain 3 items in it.
Nine (9) items will be on the board for a person to select only 3 of the items that they think is in the pre-selected envelope.
The Match and Win Promotion will be on a Friday and Saturday Night over a seven (7) week period for a total of 14 promotional nights.
Each night drawings will be conducted at 6 pm, 7 pm, 8 pm, 9 pm, 10 pm and 11 pm for a total of 6 different chances to win.
Using a combinations formula of C(n,r) there are 84 possible combinations of choosing 3 out of 9 items. (if I did that corretly)
Based on the number of combinations, there would be 84 envelopes created with each possible outcome in each envelope.
Only 6 envelopes will be selected to use each promotional night then no longer used again. So envelopes will change each night.
What is the probability of getting 3, 2, or 1 correct or nothing (0)?

Notes: Only 6 possible combinations are selected out of 84 on the first night. Then the second night will only have 78 possible combinations left and so on and so forth until the final night.

Hopefully this is easier to understand. Maybe...

Probability

Keetoowah2376

28

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Erdos

0

Why there are 84 possible outcomes? Where 84 comes from?
Schwartstack

0

I don’t understand how the different days/different drawings come into play. Are you the only person participating in these drawings? When you say “ What is the probability of getting one match out of 3?” do you mean “ever?” In any of the 42 drawings? Or on a particular drawing? Or before someone else does? Or what? It’s pretty hard to follow what you mean.
Aman R

0

Some more information regarding items and drawings needs to be added by the user.
Keetoowah2376

0

I apologize. I will edit it and see if I can explain it better. I extended the deadline.
Keetoowah2376

0

I completely revised it, so I hope it helps.
Mathe

0

.1. If multiple contestants are participating, how do we know some envelopes won't be repeated? 2.If there is only one contestant, notice there are 84 chances of winning (14 nights times 6 attempts per night), so eventually one of the envelopes has to be right. The question is not sufficiently clear.
Keetoowah2376

0

Once an envelope is used, it cannot be used again. The guest will be shown what is in the envelope to see if they won and then it is discarded. It will be different people chosen each night but a person can be selected more than once as long as it is a different night.
Mathe

0

So your question can be paraphrased as, "If a contestant plays, what is the probability that they get 1 item right or 2 or 3?
Keetoowah2376

0

Correct

Answer

The answer is accepted.

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Erdos

0

Why there are 84 possible outcomes? Where 84 comes from?
Schwartstack

0

I don’t understand how the different days/different drawings come into play. Are you the only person participating in these drawings? When you say “ What is the probability of getting one match out of 3?” do you mean “ever?” In any of the 42 drawings? Or on a particular drawing? Or before someone else does? Or what? It’s pretty hard to follow what you mean.
Aman R

0

Some more information regarding items and drawings needs to be added by the user.
Keetoowah2376

0

I apologize. I will edit it and see if I can explain it better. I extended the deadline.
Keetoowah2376

0

I completely revised it, so I hope it helps.
Mathe

0

.1. If multiple contestants are participating, how do we know some envelopes won't be repeated? 2.If there is only one contestant, notice there are 84 chances of winning (14 nights times 6 attempts per night), so eventually one of the envelopes has to be right. The question is not sufficiently clear.
Keetoowah2376

0

Once an envelope is used, it cannot be used again. The guest will be shown what is in the envelope to see if they won and then it is discarded. It will be different people chosen each night but a person can be selected more than once as long as it is a different night.
Mathe

0

So your question can be paraphrased as, "If a contestant plays, what is the probability that they get 1 item right or 2 or 3?

Answer 1

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Assuming that the participant gets one pick and does not get to change their pick after drawings.
Assuming that on any night, the participant is not aware of the results of the previous nights drawings.

For each drawing
Probability of getting all 3 correct is obviously 1/84.

Probrability of getting exactly 2 correct is 18/84
Rationale: there are 3 ways to choose the 2 correct ones out of the 3 possible ones. The remaining item has to be one of the 6 that is not in the draw. (Hence 3x6=18 possibilities.) Let say the items are digits 1-9. If the draw is 123. There are 18 ways to get exactly 2 correct: 12[4-9], 1[4-9]3, [4-9]23.

Probrability of getting exactly 1 correct is 45/84
Rationale: There are 3 ways to choose which of the 3 items is correct. For the remaining items you have to pick 2 from the 6 that were not in the draw. Hence the number of possibilites is 3x (6x5/2)= 45.

Probrability of getting exactly 0 correct is 20/84
Rationale: you have to pick 3 items out of the 6 that are not in the draw. There are (6 choose 3) ways to do this: 6x5x4/6 = 20.

Rationality check: Either you get 0 or 1 or 2 or 3 correct in each drawing so the sum of all probabilites should be one and it indeed is: (1+18+45+20)/84

For each night
Probrability of getting all 3 correct once (more is not possible because the drawings are unique) is one minus the probability of getting it wrong 6 times in a row which is 1 - 83/84 x 82/83 x ... x 78/79 = 1 - 78/84 = 6/84.
Rationale: After each drawing the number of possibilities for the next drawing goes down by 1 since drawing are unique and the number of wrong choices also goes down by 1 for the same reason.

Probrability of getting exactly 2 correct at least once is one minus the probability of getting it wrong 6 times in a row which is 1 - 66/84 x 65/83 x ... x 61/79 = 0.78
Rationale: Same as the case above.

Probrability of getting exactly 1 correct at least once is one minus the probability of getting it wrong 6 times in a row which is 1 - 39/84 x 38/83 x ... x 34/79
Rationale: Same as the case above.

Probrability of getting exactly 0 correct at least once is one minus the probability of getting it wrong 6 times in a row which is 1 - 64/84 x 63/83 x ... x 59/79
Rationale: Same as the case above.

For n nights

Extend the calculations above for 6n drawings.
Rationale: Whether the drawings happen in one night or multiple nights is immaterial.

Shadmehr

29

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