/sci/ - Science & Math

[Anonymous No.5547017]

Can someone help me with this problem? My friend is in elementary statistics and he has missed class due to an illness. How do we do this problem? I'm just trying to help. inb4 retard

[Anonymous No.5547025]

>>5547017
Binomial Distribution
That's all I'm giving you

[Anonymous No.5547044]

>hits 60% of her free throws
>hits free throws
u wut m8?

[Anonymous No.5547052]

The chance that she'll make a free throw is .6 each, so work off of that.

For her to make exactly 2 shots, her chances are .6 x .6 x .4 (chance to miss) = .144 or 14.4%

For her to make at least two shots, it's just .6 x .6 since the outcome of the last shot doesn't matter, or .36, or 36%

[Anonymous No.5547065]

>>5547052
You forgot that there are multiple ways of getting 2 shots, e.g. you could have 1 miss and then 2 hits as well.

[Anonymous No.5547072]

>>5547052
He could also miss the first and hit the last two, or miss the middle one.

[Anonymous No.5547076]

>>5547025
This

[Anonymous No.5547078]

>>5547065
Always amazing to me how convinced those posters sound like they're trying to convince a TA on an assignment.

[Anonymous No.5547083]

Draw a tree. I'm sure you know how to do this. Then, look at which outcomes fit the criteria for the answers (as mentioned before, there are multiple ways you can get two hits), and add those together. Google multiplication and addition rules for statistics. Make sure your friend understands the concept of independent events.

[Anonymous No.5547688]

>>5547052
X ~ B(3, 0.6)

p(X = 2) = 3 * 0.6^3 * 0.4

p(X <= 2) = 1 - p(X = 3)
p(X = 3) = 0.6^3
therefore p(X <= 2) = 1 - 0.6^3

There you go dibshit

[transfer student F No.5547706]

>women in sports

[Anonymous No.5547712]

a = 0.6*0.6*0.4 + 0.4*0.6*0.6 + 0.6*0.4*0.6
b = a + 0.6^3

[Anonymous No.5547715]

>>5547688 here
>>5547712 is right, I fucked up the second one and did at most 2 rather tahn at least

[Anonymous No.5547716]

>>5547688
>p(X = 2) = 3 * 0.6^3 * 0.4
... be careful with them typos: ...0.6^2 ....

also b) says "at least 2 shots", i.e. 1-p(X=1)-p(X=0) = p(X=2) + p(X=3) = 1-3*0.6*0.4^2-0.4^3 = 3*0.6^2+0.6^3

[Anonymous No.5547755]

>>5547017
okay since it is related: on a more general note, we have a binomially distributed sample with p and q=1-p as our probabilities

taken n tests:
-for exactly k hits, we say:
P(X=k) = (n over k) <span class="math"> \cdot {p}^{k} \cdot {q}^{n-k} [/spoiler]
-for e.g. at least k hits (or max n-k fails), we say:
P(X>=k) = <span class="math"> \sum_{i=k}^{n} (n \; over \; k) \cdot {p}^{k} \cdot {q}^{n-k} [/spoiler]

so my question is: are there simply ways to calculate this sum? Of course it can be replaced by 1-P(X=0) for k>=1 or alike, but if we have n=100 and k=50, do we really need to do that sum or it there a nice series?

also: how do I array?

[Anonymous No.5547757]

>>5547755
well fuck: replace k with i in the sum of course

[Anonymous No.5547965]

>>5547755
noone?

[Anonymous No.5547994]

>>5547017
Jesus Christ anyone who can't solve this should just drop the class

[Anonymous No.5548063]

>>5547755
>how do I array?
like this maybe
<div class="math">\left.\begin{pmatrix}
n \\
k \right</div>

[Anonymous No.5548071]

/sci/, you are a disgrace.
Why do you help this dumbass with his homework?

[Anonymous No.5548081]

>>5548071
because that's the maximum mental capacity of /sci/
anything harder and no one replies

[Anonymous No.5548080]

>>5548071
/sci/ is irremediably shit anyway, /sci/ itself has become a disgrace so who cares if some help this nigglet with his homework

[Anonymous No.5548083]

>>5548063
very helpful, letting me read the page source ....

<span class="math"> \begin{pmatrix} n \\ k \end{pmatrix} [/spoiler]
just testing this