Mr. and Mrs. X (1 Viewer)

[Dreamerish*~]

From a group of 9 people, including Mr. and Mrs. X, a committee of 5 people is to be chosen. Mr. X will not join the committee without his wife, but his wife will join the committee without her husband (LOL). In how many ways can the committee be formed? Find the probability that a particular committee does not include both Mr. and Mrs. X.

For the first bit, since Mr. X will not join the committee without his wife, then the possibilities are:

Both Mr. and Mrs. X are in: ⁷C₃ = 35
Neither Mr. nor Mrs. X are in: ⁷C₅ = 21
Only Mrs. X is in: ⁷C₃ = 35

. : Total number of ways the committee can be arranged = 91.

However, for the second part, the number of arrangements would be 91 - 35 = 56.

So the probability that it does not include both Mr. and Mrs. X is 56/91 = 8/13, right?

But the answer has it as 4/9, which is 56/126, 126 being ⁹C₅. I'm just thinking, since they've already told us that Mr. X can't be in it without his wife, how can the total number of arrangements be ⁹C₅.

Sounds like a majorly stupid question, I know, but someone help me?

 

[KFunk]

Your reasoning sounds solid, whoever wrote the answer is just being a gronk.

 

[insert-username]

I think the answer's wrong as well. Your reasoning works out, so in all likeliness it's probably a misprint in the answers.


I_F

 

[damo676767]

no the ansewer is right, a commity weather it can formed or not is still a posibility

 

[KFunk]

How can it be considered a possibility if the probability of it occuring = 0?

 

[damo676767]

it is not, if i pic 5 people at random there is a chance that i will pick him and knot her

 

[Dreamerish*~]

it is not, if i pic 5 people at random there is a chance that i will pick him and knot her

But in this question, it was clearly stated that he is not to join the committee without her, so that rules out him being in it alone as a possibility.

 

[KFunk]

But the way the question is phrased suggests that that situation is not a possibility. Alternatively I geuss you could say that the people picking the comitee members are unaware of Mr X's issues so they might inadvertently create that conflicting situation but then doesn't that just disregard the conditions in the question?

 

[damo676767]

i think we are unlikley to get a quesyion that is as ambigious as that

and if we did both would be correct

 

[shivi]

Over 126 not 91

Probabitily of (neither + Ms X alone) / (Total possibilities)
(7C4 + 7C5) / 9C5 = 4/9

 

[Riviet]

I know you have already completed the HSC and got your results long ago, but I thought this question was interesting... so yeah, bump!

You still need to include the chance of both of them being in the committee because probability is calculated by desired number of outcomes divided by total number of outcomes. So don't subtract from the absolute total number of outcomes when finding the probability.