What percent make positive numbers?
- Thread starter Loki123
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Correction, it's supposed to say - 3
BigBeachBanana
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How much do you know about continuous probability density and cumulative distribution functions.
[math]\Pr(X \le a)= \int_{\-\infty}^{a} f(x) dx[/math]
PS: I didn’t get a chance to look at your other question.
Nothing right now. But I have probability classes this week so maybe I'll know something. Although from what I know thr questions are mostly like
You have 5 balls, 3 are white, 2 are yellow, one is black, how many do you have to take (not looking) to surely have a black ball.
BigBeachBanana
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You’re talking about discreet probability. Whereas this question talks about an interval, which deal with continuous probability,require integration.
pka
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Three white balls, two yellow & one black make six balls not five.
BigBeachBanana
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@Loki123, since you haven't learned anything related to this topic. I don't mind providing a solution.
The answer seems to assume that the probability density is uniform, meaning you have equally likely chance to pick a number within the interval.
Let [imath]X[/imath] be a random variable represents the number that get pick out of the interval. So the probability of picking any number with in the interval is: [imath]p(x)=\frac{1}{b-a}=\frac{1}{2-(-3)}=\frac{1}{5}[/imath].
Now, to find the probability of picking a positive number, this means we're picking [imath]x[/imath] from the interval [imath](0,2)[/imath].
Lastly, sum up all the probability of picking all the numbers within [imath](0,2)[/imath].
[math]\Pr( 0 < X <2 ) =\int_{0}^{2}\frac{1}{5}\, dx=0.4[/math]
The answer seems to assume that the probability density is uniform, meaning you have equally likely chance to pick a number within the interval.
Let [imath]X[/imath] be a random variable represents the number that get pick out of the interval. So the probability of picking any number with in the interval is: [imath]p(x)=\frac{1}{b-a}=\frac{1}{2-(-3)}=\frac{1}{5}[/imath].
Now, to find the probability of picking a positive number, this means we're picking [imath]x[/imath] from the interval [imath](0,2)[/imath].
Lastly, sum up all the probability of picking all the numbers within [imath](0,2)[/imath].
[math]\Pr( 0 < X <2 ) =\int_{0}^{2}\frac{1}{5}\, dx=0.4[/math]
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discrete probability
"Discreet" involves being careful about one's speech or actions.
Where did this come from
Does it not make a difference that 2 isn't included unlike the other numbers?
BigBeachBanana
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It comes from the continuous uniform distribution. You can learn more about it here.
For continuous probability, the inclusion does not impact the result because, in general:
[math]\Pr(X \le x)=\Pr ( X < x ) + \Pr( X = x )=\int_{a}^{x}f(x)\,dx+\int_{x}^{x}f(x)\,dx=\Pr( X < x )+0\\ \therefore \Pr(X \le x)=\Pr( X < x )[/math]
In simple terms, the probability of picking an exact number within a continuous interval is 0 as there are infinitely many. However, this is not the case for discrete probability.
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