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/sci/ - Science & Math

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2494964 No.2494964 [Reply] [Original]

Hey could you guys help me with a few stat problems (2)?

A senate senator believes that 25% of all senators on the Finance Committee with strongly support the tax proposal she wishes to advance. Suppose that this belief IS CORRECT and that five senators are approached at random. What is the probability that at least one of the five will strongly support the proposal? What is the probability that a majority of the five will strongly support the proposal?

>> No.2494991 [DELETED] 

Anyone :| please?

>> No.2495045

bump

>> No.2495064

Problem one: 1 - (3/4)^5

Problem two: do your own HW

>> No.2495080

Wait what

How?

I just don't get this shit at all T.T

>> No.2495081

Parameter is known.
Sample population of 5 is taken.
First answer is inverse chance of failure, so 1-[5*(.25)]=.25

Second answer requires you to calculate several permutations, i.e. chances of 5 supporters, chances of 4 supporters 1 other, chance of 3 supporters 2 other, etc.

>> No.2495089

Also, that would make the probability .237

How is that possible? That's less than 25% in a larger pool.. Shouldn't the probability be at least higher than .25 in a group of 5 or more people?

>> No.2495098

>>2495081

Continuing, to get the answer to the second one, you add the probabilities of the successful permutations together.

[5s,0f]+[4s,1f]+[3s,2f]

>> No.2495099

>>2495081
Okay, so the guy above is wrong?

thanks btw

>> No.2495115

>>2495098
I don't know how to do that

Even my book is gibberish to me.

I don't understand why I'm so bad at stats, I had no trouble with math up to Multivariable Calc, but then this stuff chokes me up every time.

>> No.2495117

>>2494964

Use binomial distribution. i.e
your random variable (number of senators supporting) X ~ bin(n=5, p= 0.25)

You want P(X=1)
and P(X>2) = P(X=3) + P(X=4) +P(X=5) =

http://en.wikipedia.org/wiki/Binomial_distribution
http://stattrek.com/Tables/Binomial.aspx

>> No.2495136

>>2495117
oh shit that n and p stuff rings a bill

I have a few equations here in my notes

Okay so to figure about probability would I use the PMF equation?

P(x) = (n!/x!(n-x)!)P^x(1-p)^(n-x)

?

>> No.2495146

>>2495115

Don't sweat it mate. Probability can be incredibly counter-intuitive and hard to follow.

Once you get the concept of a random variable and particular distributions its all a matter of interpreting the question and deciding which distribution you're working with.

Having said that I admit I'm not great with combinatorial style probability.

>> No.2495171

>A senate senator believes that 25% of all senators on the Finance Committee with strongly support the tax proposal she wishes to advance.
>five senators are approached at random

What is the probability that a senator approached at random will be on the finance committee?

>> No.2495179

>>2495136

Okay so in the first question you want;
P(X=1), so put all your values in for that...
x=1, p=0.25 (hence 1-p = 0.75), n=5

just plug them into the binomial formula (pmf this case, we are only looking at ONE scenario, x=1)

For part two you would use the CUMULATIVE function as we're interested in an entire tail for the distribution.

>> No.2495188

>>2495179
Yup got it thanks

Okay now, what about the cumulative thing you're talking about?

You mean just get the PMF again for x = 3 4 and 5? and then add up the probabilities?

Because that's what I was thinking

>> No.2495200

>>2495188
Sounds right to me.

There are some other tricks you could use to make it much faster.

Take for instance.. working out P(X=2)..
Note: P(X>2) = 1- P(X<3) ..
and P(X<3) = P(X=1) + P(X=2)

Either way you will get the same answer as the method you outlined in ;
>>2495188

>> No.2495356

Sorry, but I have one more question guys

But I just need to know whether it's hyper geometric or binomial etc and i can figure out the rest myself

A committee of eight members is to be formed from a group of eight men and eight women. If the
choice of committee members is made randomly, what is the probability that precisely half of these
members will be women?

>> No.2496209
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2496209

Probability totals 100%. When a problem calls for AT LEAST one, treat it like:

100% minus the probability that zero senators in your sample are in support.

The probability that each senator doesn't support it is 75%. The probability that NO senators in your sample supports it is:
.75*.75*.75*.75*.75 = 0.237 or 24%, rounding up.

The probability that AT LEAST ONE senator supports it, therefore, is 100-24 = 76%

>> No.2496230

(.25^5) + (.25^4 x .75) + (.25^3 x .75^2)

Second part=

>> No.2496378

Can't solve. Need to know how many senators are on the Finance Committee first.

Stupid question with no solution.

>> No.2496387

binomial distribution bro, thank me later.

>> No.2496390

>>2495179
requiring a majority doesn't entail a cumulative distribution. It just wants to know the probability of 3 saying yes.

>> No.2496394

>>2496378
>stupid question with no solution
I can't maths, sympathize pl0x
inb4 gtfo my /sci/

>> No.2496421

>>2496394
I could solve it if it could be solved. I could give you the answer the question is LOOKING for, but it would be wrong.

It needs to be figured out first how many senators on the finance committee there are, what the percentage of people supporting it there are, and from there it's possible to know the probability of at least one of person strongly supporting it, and from there the probability that 3 or more would support it.

It's not possible to know without knowing how many senators are on the finance committee because you're working with incomplete information. It tells you that 25% of all of the senators would support it, but you don't know how many that is. It could be 5, making it impossible for a majority of the senators to support it. It could be 20, and have a 25% chance of them all supporting it. You don't know.

It's not possible to figure out without all of the information. Any answer you could give would be wrong.

>> No.2496427
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2496427

>>2495356
The probability of picking the first member is 0.5. 8 men, 8 women, you want a woman.

Since you can't pick a woman twice (because women are beetches), the probability of picking a second woman is 7/15.

The probability of picking EXACTLY 4 women and 4 men randomly is:
8/16 * 7/15 * 6/14 * 5/13 * 8/12 (8 men, 12 left) * 7/11 * 6/10 * 5/9
= 0.5 * 0.46 * 0.43 * 0.38 * 0.66 * 0.63 * 0.6 * 0.55
= 0.005
or 0.5%

Any more problems, or can I go j/o now?

>> No.2496459

The probability of having a majority is given by:
(0.25^3)(1 - 0.25 * 0.75)