Hard math question... help?

Question

There are 6 married couples attending a party where two door prizes (the prizes are the same) are awarded to two different people. Solve to 3 decimal places.

What is the probability a married couple wins the prizes?

What is the probability one prize goes to a man and one to a woman?

If Mary does not win a prize, then what is the probability that John will?

Answer 1

There are sixty-six possible outcomes in the first two scenarios, and fifty-five possible outcomes in the third (Mary does not win):

What is the probability a married couple wins the prizes?
9.09% (6/66 x 100)

What is the probability one prize goes to a man and one to a woman?
54.5% (36/66 x 100)

If Mary does not win a prize, then what is the probability that John will?
18.2% (10/55 x 100)

(or 0% if John never made it to the party!)

Answer 2

EDIT: The ONLY person to get all 3 right answers is the answerer below me. I got the first 2 parts, and so did Wal C. I misinterpreted the last part as "Find the probability that Mary does not get a prize and John does"(I'm not even sure I did THAT correctly). (I haven't corrected it so you can see my mistake). Congratulations, Tak!

(These are based on the assumption that after a person gets the first prize, they are not eligible for the 2nd)

a)
1. Prob. that someone gets the 1st door prize = 1
2. Prob. that this person's spouse gets the 2nd prize = ?
Let's see, there are 11 people left to get the 2nd door prize, and only 1 of them is the first person's spouse...so the probabity of this event is 1/11.
Now multiply the 2 together to get the overall probability:
P = (1)(1/11) = 1/11

b)
1. Prob. that someone gets the 1st door prize = 1
2. Prob. that the 2nd prize goes to someone of the opposite sex = ?
Ok, there are 11 people left and 6 are of the opposite sex...so the probability of this event is 6/11.
So the overall probability of this situation is:
P = (1)(6/11) = 6/11

c)
This one's a little tricky. Take it in 2 parts.
1. Prob. that Mary doesn't get a prize = ?
Prob. that she gets the 1st prize = 1/12
Prob. that she does NOT get the 1st prize = 11/12
Prob. that she gets the 1st and gets the 2nd = 0
Prob. that she gets the 1st and does NOT get the 2nd = 1/12
Prob. that she does NOT get the 1st and gets the 2nd = 1/12
Prob. that she does NOT get the 1st and does NOT get the 2nd
= (11/12)(10/11) = 10/12 = 5/6
All YOU care about is the last option (she gets no prize).
2. Prob. that John gets a prize = ?
Go through the same pricess as before:
Prob. that he gets the 1st prize = 1/12
Prob. that he does NOT get the 1st prize = 11/12
Prob. that he gets the 1st and gets the 2nd = 0
Prob. that he gets the 1st and does NOT get the 2nd = 1/12
Prob. that he does NOT get the 1st and gets the 2nd = 1/12
Prob. that he does NOT get the 1st and does NOT get the 2nd
= 10/12
YOU are interested in the probability that he gets a prize, which is:
[Prob. that he gets the 1st prize] + [Prob. that he does NOT get the 1st and gets the 2nd] = (1/12) + (1/12) = 2/12 = 1/6
Now, multiply the prob. from part 1 by the prob. in part 2:
P = (5/6)(1/6) = 5/36

Answer 3

i. What is the probability a married couple wins the prizes?

Well someone wins the first prize.
Only 1 person in the remaining attendees will be that partner of the winner.
So the chances that that person wins the other prize is 1/11 (≈ 0.091)

ii. What is the probability one prize goes to a man and one to a woman?

Well someone wins the first prize.
Then there are 6 people of the opposite sex (assuming that the couples are heterosexual of course!!)
So the probability that one of them wins the second prize = 6/11 (≈ 0.545)

iii. If Mary does not win a prize, then what is the probability that John will?

Someone wins first prize

If it is not Mary then P(not Mary) = 11/12
Then P(John) = 1/11
So P(not Mary and THEN John) = 11/12 * 1/11
= 1/12
BUT there is also P( John and then NOT Mary) and this also is 1/12

So P(Mary does not win and John does) = 2 * 1/12
= 1/6
(≈ 0.167)

And of course this excludes the possibility of one person winning BOTH prizes

If this were possible then the probabilities change to:
i. 1/12 (≈ 0.083)
ii. 6/12 (= 0.5)
iii. 2 * 11/12 * 1/12 = 11/72 (≈ 0.153)

Answer 4

There are 12 people altogether, meaning each couple has two chances at winning each prize.

1) If the married couple wins both prizes we must multiply the probabilities of each of them winning.

To win the first prize, you have a 1/12 chance and to win the second one, you have a 1/11 chance (because we assume the same person can't win both prizes). That means, for one couple to win both prizes there is a 1/121 chance, or 0.826% chance of winning.

2) The odds of one prize going to a man and one going to a woman is as follows.

For a man to win one prize, he has a 1/12 chance, but because there are 6 men, there is a 1/2 chance that a man wins. The odds of the woman winning are the same. Multiply both results 1/2 * 1/2 = 1/4 = 25% chance of a man and woman winning the prize.

3) Mary has a 1/12 chance of winning the prize, but if she does not, that means John should have a 1/11 chance or 9.091% chance of winning.

Hopefully these are right (I'm much better at calculus :D)

Answer 5

silvcubo is right the probability a married couple wins the prizes is 1/6 the probability one prize goes to a man and one to a woman 1/6 the probability that John will 1/11

Answer 6

1) 1/11 = 0.909 Basically, there are 12 ppl in total, it wouldn't matter who you pick to win for the first prize. For the second prize, u only need to match to the spouse of the person who win the first prize. And there are 1 out of 11 chances. (since the first person has to be out of the picture, can't win two prizes)


2) 6/11 = 0.545. Again the first person who win the first prize does not matter. Of the remaining 11 ppl, there will be 6 who are opposite sex.

3) 2/11 = 0.182 We are given that Mary does not win a prize, so we can eliminate her already. (Conditional probability). so there are 11 ppl competitng for the two prizes. The probability that John does not win is 10/11 * 9/10 = 9/11, which means the probability that he will win is 2/11 = 0.182

Answer 7

The married couple wins the prizes:
since order does not matter the first prize given away can go to either of the people in the couple, so probabilty is 2/12, and since the second prize would have to go to the other person the probablilty would be 1/11. Since it is an "AND" question you would multiply them, so (2/12)(1/11) = 2/132, and since there are 6 couples we would multiply that by 6, so 12/132.

A man and a woman - You could have a man and woman, or, woman and a man. "AND" means multiply and "OR" ,means add. So : (6/12)(6/11) + (6/12)(6/11) = (36/132) + (36/132) = 72/132

Mary no AND john yes: Mary no = (11/12)(10/11)=110/132
John yes= (1/12)+(1/11)=23/132

Now multiply them: (110/132)(23/132)= your ans (I do not have a calc in front of me)

Answer 8

well the chances for a couple are a .33%(2 out of 6). A 0.1666666666667% cahnce a at least 1 man or woman getting a prize. And .20% that John will get a prize (1 out of 5)

Answer 9

a million+a million=2 besides, somebody with far too plenty time on their palms desperate to instruct it... "The evidence starts off from the Peano Postulates, which define the organic numbers N. N is the smallest set pleasing those postulates: P1. a million is in N. P2. If x is in N, then its "successor" x' is in N. P3. there is not any x such that x' = a million. P4. If x isn't a million, then there's a y in N such that y' = x. P5. If S is a subset of N, a million is in S, and the implication (x in S => x' in S) holds, then S = N. then you definately ought to define addition recursively: Def: enable a and b be in N. If b = a million, then define a + b = a' (utilising P1 and P2). If b isn't a million, then enable c' = b, with c in N (utilising P4), and define a + b = (a + c)'. then you definately ought to define 2: Def: 2 = a million' 2 is in N by utilising P1, P2, and the definition of two. Theorem: a million + a million = 2 evidence: Use the 1st area of the definition of + with a = b = a million. Then a million + a million = a million' = 2 Q.E.D. word: there is an option formula of the Peano Postulates which replaces a million with 0 in P1, P3, P4, and P5. then you definately ought to alter the definition of addition to this: Def: enable a and b be in N. If b = 0, then define a + b = a. If b isn't 0, then enable c' = b, with c in N, and define a + b = (a + c)'. you besides mght ought to define a million = 0', and 2 = a million'. Then the evidence of the thought above is a sprint distinctive: evidence: Use the 2d area of the definition of + first: a million + a million = (a million + 0)' Now use the 1st area of the definition of + on the sum in parentheses: a million + a million = (a million)' = a million' = 2 Q.E.D." Wow, he could be a real hit with the ladies...!! :)

Answer 10

to the 1st question, 0.694%

2nd. 25%

3rd. 9%

i think....


you guys are getting it wrong because the prize doesnt go to a couple, it goes to a person.. 12 people, two prizes