Answer:
0.0111% probability that he answers at least 10 questions correctly
Step-by-step explanation:
For each question, there are only two outcomes. Either it is answered correctly, or it is not. The probability of a question being answered correctly is independent from other questions. So we use the binomial probability distribution to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
![P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20C_%7Bn%2Cx%7D.p%5E%7Bx%7D.%281-p%29%5E%7Bn-x%7D)
In which
is the number of different combinations of x objects from a set of n elements, given by the following formula.
![C_{n,x} = \frac{n!}{x!(n-x)!}](https://tex.z-dn.net/?f=C_%7Bn%2Cx%7D%20%3D%20%5Cfrac%7Bn%21%7D%7Bx%21%28n-x%29%21%7D)
And p is the probability of X happening.
A multiple-choice examination has 15 questions, each with five answers, only one of which is correct.
This means that ![n = 15, p = \frac{1}{5} = 0.2](https://tex.z-dn.net/?f=n%20%3D%2015%2C%20p%20%3D%20%5Cfrac%7B1%7D%7B5%7D%20%3D%200.2)
What is the probability that he answers at least 10 questions correctly?
![P(X \geq 10) = P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15)](https://tex.z-dn.net/?f=P%28X%20%5Cgeq%2010%29%20%3D%20P%28X%20%3D%2010%29%20%2B%20P%28X%20%3D%2011%29%20%2B%20P%28X%20%3D%2012%29%20%2B%20P%28X%20%3D%2013%29%20%2B%20P%28X%20%3D%2014%29%20%2B%20P%28X%20%3D%2015%29)
![P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20C_%7Bn%2Cx%7D.p%5E%7Bx%7D.%281-p%29%5E%7Bn-x%7D)
![P(X = 10) = C_{15,10}.(0.2)^{10}.(0.8)^{5} = 0.0001](https://tex.z-dn.net/?f=P%28X%20%3D%2010%29%20%3D%20C_%7B15%2C10%7D.%280.2%29%5E%7B10%7D.%280.8%29%5E%7B5%7D%20%3D%200.0001)
![P(X = 11) = C_{15,11}.(0.2)^{11}.(0.8)^{4} = 0.000011](https://tex.z-dn.net/?f=P%28X%20%3D%2011%29%20%3D%20C_%7B15%2C11%7D.%280.2%29%5E%7B11%7D.%280.8%29%5E%7B4%7D%20%3D%200.000011)
![P(X = 12) = C_{15,12}.(0.2)^{12}.(0.8)^{3} \cong 0](https://tex.z-dn.net/?f=P%28X%20%3D%2012%29%20%3D%20C_%7B15%2C12%7D.%280.2%29%5E%7B12%7D.%280.8%29%5E%7B3%7D%20%5Ccong%200%20)
![P(X = 13) = C_{15,13}.(0.2)^{13}.(0.8)^{2} \cong 0](https://tex.z-dn.net/?f=P%28X%20%3D%2013%29%20%3D%20C_%7B15%2C13%7D.%280.2%29%5E%7B13%7D.%280.8%29%5E%7B2%7D%20%5Ccong%200%20)
![P(X = 14) = C_{15,14}.(0.2)^{14}.(0.8)^{1} \cong 0](https://tex.z-dn.net/?f=P%28X%20%3D%2014%29%20%3D%20C_%7B15%2C14%7D.%280.2%29%5E%7B14%7D.%280.8%29%5E%7B1%7D%20%5Ccong%200%20)
![P(X = 15) = C_{15,15}.(0.2)^{15}.(0.8)^{0} \cong 0](https://tex.z-dn.net/?f=P%28X%20%3D%2015%29%20%3D%20C_%7B15%2C15%7D.%280.2%29%5E%7B15%7D.%280.8%29%5E%7B0%7D%20%5Ccong%200%20)
![P(X \geq 10) = P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15) = 0.0001 + 0.000011 = 0.000111](https://tex.z-dn.net/?f=P%28X%20%5Cgeq%2010%29%20%3D%20P%28X%20%3D%2010%29%20%2B%20P%28X%20%3D%2011%29%20%2B%20P%28X%20%3D%2012%29%20%2B%20P%28X%20%3D%2013%29%20%2B%20P%28X%20%3D%2014%29%20%2B%20P%28X%20%3D%2015%29%20%3D%200.0001%20%2B%200.000011%20%3D%200.000111)
0.0111% probability that he answers at least 10 questions correctly
Answer:
Step-by-step explanation:
The formula is y = mx + b
m being the slope, rise over run. And b being the y-intercept. Right off the bat we can visually see the y-intercept is -4.
To find slope, we need to take two sets of coords and apply the slope fomula. The slope fomula is change in y divided by the change in x. The function itself is straight, so that means the slope will be the exact same no matter which points you choose.
(4, -1) and (8, 2) are coords on the line. Do 2 - (-1) to get 3. then do 8 - 4 to get 4. Finally, we just gotta do 3/4 which is simply
.
We have the slope of 3/4 and we have the y-intercept of -4. Just plug it in the standard formula of y = mx + b to get:
![y=\frac{3}{4} x-4](https://tex.z-dn.net/?f=y%3D%5Cfrac%7B3%7D%7B4%7D%20x-4)
Answer:
A
Step-by-step explanation:
Choice A is right because there is no dot above the two.
Choice B is not right because there is 2 dots above 1 and not 3.
Choice C is not right because the data doesn't go passed 5 to get to 8
Choice D is not right because it was 5 games because there is 5 dots above 3