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# Sam and Jessica are invited to a dance. If there are 7 men

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Joined: 31 Dec 1969
Location: Russian Federation
WE: Supply Chain Management (Energy and Utilities)
Sam and Jessica are invited to a dance. If there are 7 men  [#permalink]

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25 Jun 2004, 13:05
Sam and Jessica are invited to a dance. If there are 7 men and 7 women in total at the dance, and one woman and one man ar chosen to lead the dance, what is the probability that Sam and Jessica will NOT be chosen to lead the dance?

I have a basic probability question. In the question above, the answer is 48/49 (i.e. 1-(1/49)). I understand the rationale of this (1 minus the prob. that they will be chosen), but I originally solved the problem as 36/49 (i.e. 6/7 * 6/7 - the probability that they each won't be picked, as the prob. that they won't be picked is 6/7 each). Why can't I solve it like this? This problem comes up on dice problems too, so I would like to understand the concept. Here's another problem that emcompasses the same concept:

A six-sided die with faces 1 to 6 is rolled three times. What is the prob. that the face with the number 6 on it will NOT be facing upwards on all three rolls?

Conceptually, what I am missing that I can't solve it as 5/6 * 5/6 * 5/6? (correct answer is 215/216)

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Intern
Joined: 04 Aug 2003
Posts: 29
Location: Texas

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25 Jun 2004, 15:17
1
The first problem asks for P(Sam AND Jessica is not chosen) . So we should not consider the situation where Either Sam or Jessica is not chosen.

Second problem might be a bit complex.

The problem asks for No sixes must face up in ALL 3 rolls.

ie, it is okay for having 1 or 2 sixes in 3 rolls. In your solution you have not considered the chances for 1 or 2 Sixes rather you have only considered no in all 3 rolls

Let N be no six in a roll and S be a Six in the roll

The following are the acceptable distributions.

N N N = 5/6*5/6*5/6 = 125/216
N N S = 5/6*5/6*1/6 = 25/216
N S N = 5/6*5/6*1/6 = 25/216
S N N = 5/6*5/6*1/6 = 25/216
N S S = 5/6*1/6*1/6 = 5/216
S N S = 5/6*1/6*1/6 = 5/216
S S N = 5/6*1/6*1/6 = 5/216

Adding all the above we get 215/216. But you have considered only the first option that gave you 125/216

The easiest approach is to use the complemet, where we consider only
S S S and take the complement of it

ie, 1/6*1/6*1/6 = 1/216 Therefore ans = 1-1/216 = 215/216
Senior Manager
Joined: 21 Mar 2004
Posts: 405
Location: Cary,NC

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26 Jun 2004, 22:25
Using Probability notation theory one might be able to understand or explain your doubt.
Let a and b be the two events. a&b is the two events happening together
not P(a) is the probability of the event a not happening.

P( a & b) = P(a).P(b) -------- This is the rule

not P(a&b) = 1- P(a).P(b)

what you did was calculate the below for the answer :

This is because :

not P(a&b) is NOT = not P(a). not P(b)

- HTH and clarifies your doubt
- ash
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ash
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Intern
Joined: 09 May 2004
Posts: 27
Location: houston, tx

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28 Jun 2004, 10:02
Hi guys,

Thanks so much for the explanations (that was me who posted earlier - forgot to log in). I completely understand why my logic was off before. Essentially what you guys are saying is that if I want to calculate the probability that events A AND B do NOT happen, it is not sufficient to multiply not P(a) * not P(b) because this only considers the specific probability that A and B both do not happen, versus the other possibilities where both A AND B may not happen.

But what if we changed the question around to this:

A six-sided die with faces 1 to 6 is rolled three times. What is the prob. that the face with the number 6 on it will be facing upwards exactly twice?

Would it be: Prob of events * possible combinations of having a six exactly twice?
= (1/6 * 1/6 * 1/6) * (3! / ( 2!/1!))
= (1/216) * 3
= 3/216

Senior Manager
Joined: 21 Mar 2004
Posts: 405
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28 Jun 2004, 10:16
Here's my solution.

There are 3 specific cases where 6 comes twice exactly.

A = 6,6,N
B = 6,N,6
C = N,6,6

N = 6 doesnt come on that throw

P(A) = 1/6 * 1/6 * not P(6) = 1/36 * ( 5/6) = 5/216

P(B) and P(C) are the same as P(A).

now P(A or B or C ) = P(A) + P(B) + P(C) = 15/216

- ash
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ash
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I'm crossing the bridge.........
Intern
Joined: 09 May 2004
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28 Jun 2004, 14:53
Ok, I gotcha, thanks. I calculated the probability wrong at first, but I understand now. And if they asked "at least" twice instead of "exactly" twice, how would you solve it?

I think you would you add 1/216 (i.e. 1/6 * 1/6 * 1/6 - prob. of rolling "6" all three times) to the probability you got above... but would you have to do that 3 times to account for the 3 possibilities you could roll "6" all three times - or just once?
Senior Manager
Joined: 21 Mar 2004
Posts: 405
Location: Cary,NC

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29 Jun 2004, 09:25
at least twice ......would bring the following cases into picture. ( It would be an extension of the previous case)

A = 6,6,N
B = 6,N,6
C = N,6,6
D = 6,6,6 ( the extra event)

P(A) = P(B) = P(C) = 5/216

P(D) = (1/6)*(1/6)*(1/6) = 1/216

now P(A or B or C or D ) = P(A) + P(B) + P(C) + P(D)= 6/216 = 1/36

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ash
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I'm crossing the bridge.........
Intern
Joined: 09 May 2004
Posts: 27
Location: houston, tx

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29 Jun 2004, 18:21
Great, thanks. Makes perfect sense now....
Intern
Joined: 09 May 2004
Posts: 27
Location: houston, tx

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05 Jul 2004, 10:34
Hi ashkg,

I was reviewing this post, and I understand the concept you were explaning, but I think you may have added wrong for the problem above? When you say:

"now P(A or B or C or D ) = P(A) + P(B) + P(C) + P(D)= 6/216 = 1/36"

Don't you mean to add 1/16 (Probability of D) to the 15/216 (sum of probability of A, B, C), which would yield a total probability of 16/216? Because the probability would only get larger if you add more possibilities that you could roll it, correct? (i.e. "twice" to "at least twice")

Like I said, I understand the concept, but just wanted to clarify with you in case I am missing something or calculating something wrong.

Thanks!
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Re: Sam and Jessica are invited to a dance. If there are 7 men  [#permalink]

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04 Jul 2017, 06:49
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Re: Sam and Jessica are invited to a dance. If there are 7 men   [#permalink] 04 Jul 2017, 06:49
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