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/sci/ - Science & Math

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5033720 No.5033720 [Reply] [Original]

Hey, /sci/

A local company has two job openings available, and 170 people apply. Of the applicants, 109 are men and 61 are women. If two people are selected at random, what is the probability that exactly one is a woman? Please explain your methods/logic. Thanks!

>> No.5033741

bump

>> No.5033771

100% because sexism.

>> No.5033810

shameless self bump

>> No.5033818

50%, either she is or he isn't.

>> No.5033831

>>5033818
could be a tranny....

>> No.5033836

Just make a tree, jackass.

>> No.5033849

Well, there are four possibilities:
A is a man, B is a man
A is a man, B is a woman
A is a woman, B is a man
A is a woman, B is a woman

The second and third outcomes each feature exactly one woman. That's two of four outcomes, or 50%.

>> No.5033853

50%

>> No.5033859

>>5033849

I really can't tell if this one is trolling or retarded.

>> No.5033864

you idiots, there are 3 options
0 women
1 woman
2 women
33% chance

>> No.5033877

>>5033859
Well, there are three possibilities:
I am lacking a chromosome.
I have the correct number of chromosomes.
I have a bonus chromosome.

67% chance that I'm retarded.

>> No.5033884

>>5033864
It could be either 50% or 33%. It depends on whether the birth order of the women matters. It's an ambiguous troll question.

>> No.5033887

Depends... are these two people being selected from a kitchen?

>> No.5033895

>>5033864
Nope, it's 25%. There are two ways you could have 1 woman. She could be chosen first or last.

>> No.5033891

>>5033887
screw you idiot

>> No.5033896

chance of 1st being woman and 2nd man
+
chance of 1st being man and 2nd woman

so 61/170 * 109/169
+
109/170 * 61/169

but yeah im probably being trolled

>> No.5033901

The probability of 2 men = 109/170 * 108/169.
The probability of 2 women = 61/170 * 60/169

Probability of exactly one woman = 6649/14365

>> No.5033903

Think of it like a coin with 109 (64.11%) chance for heads and 61 (35.88%) chance for tails, what are the chances that you flip twice and and get one heads and one tails?

That's all the homework I'll do for you, OP

>> No.5033927 [DELETED] 

>>5033903

This. See Binomial Distribution. It's all about probabilities regarding success and failure

>> No.5033980

>>5033859
Just like how when you flip coins, it's just as likely that you'll get two head as it is that you'll get heads and then tails. Did you ever study entropy?

If I have two chambers which form a closed system TOGETHER with 50 particles among them, is it equally likely that I'll have 25 in each as it is that I'll have all 50 in one chamber? think about that for a second, and perhaps even use reality to check your answers.

>> No.5033981

>>5033980
not head and then tails, but rather head and tails once each

>> No.5033986

>>5033980
basically I can select a particle from each chamber when they have 25 each and swap them to get a unique scenario

I can do the same thing with men and women or heads and tails

>> No.5033991

>>5033980
http://archive.installgentoo.net/sci/thread/5033976

> Too retarded for basic probability
> Too retarded to post correctly

>> No.5033997

>>5033991
>ad hominem
please explain why I'm wrong, I'd really like to learn something new, even if I discover that I'm retarded in the process

>> No.5034004

>>5033997
Not an ad hominem. I wasn't arguing with you, just making fun of you.

>> No.5034006

>>5034004
I tried to post in an expanded thread. I forget that I'm not actually in the thread pretty often.

>> No.5034018

0% since there's no way my company would hire women aside from being cumdumpsters

>> No.5034049 [DELETED] 

<span class="math">P(Woman_1) = \frac{61}{170}[/spoiler]
<span class="math">P(Woman_2) = \frac{60){169}[/spoiler]

Therefore,

<span class="math">P(Woman_1 or Woman_2) = P(Woman_1) + P(Woman_2) - P(Woman_1 and Woman_2)[/spoiler]

Since

<span class="math">P(Woman_1 and Woman_2) = P(Woman_1)P(Woman_2)[/spoiler]

It can be shown that

<span class="math">P(Woman_1 or Woman_2) = P(Woman_1) + P(Woman_2) - P(Woman_1)P(Woman_2) = \frac{61}{170} + \frac{60}{169} - \frac{61}{170}\fraq{60}{169}[/spoiler]

>> No.5034051

>>5034018
Moldy Basement Inc.

>> No.5034057

>>5034051
I lol'd.

haha 9/10

>> No.5034074

I believe it's (109/160)(108/159)(61/158). Basically after every guess the number of people (denominator) decreases, so does the number of men if you choose a man.

>> No.5034092

>>5034074
It never says the same person can't be chosen twice

>> No.5034097
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5034097

35.9%

>> No.5034109

P(Woman_1 or Woman_2) = P(Woman_1) + P(Woman_2) - P(Woman_1 and Woman_2)
=> P(Woman_1 or Woman_2) = P(Woman_1) + P(Woman_2) - P(Woman_1)P(Woman_2)

You know all those values ^^

>> No.5034122

>>5034109

This is actually the right answer. The Binomial idea does not work because the probability is a function of your trial number (conditonal probability)

>> No.5034135

>>5033997
>ad hominem
>4chan
Don't tell me you're surprised.