1) From the total of 17 marbles, Jacob takes 2 out at random. There are
![\dbinom{17}2=C(17,2)=C^{17}_2=\dfrac{P(17,2)}{2!}=\dfrac{P^{17}_2}{2!}=\dfrac{17!}{2!(17-2)!}=136](https://tex.z-dn.net/?f=%5Cdbinom%7B17%7D2%3DC%2817%2C2%29%3DC%5E%7B17%7D_2%3D%5Cdfrac%7BP%2817%2C2%29%7D%7B2%21%7D%3D%5Cdfrac%7BP%5E%7B17%7D_2%7D%7B2%21%7D%3D%5Cdfrac%7B17%21%7D%7B2%21%2817-2%29%21%7D%3D136)
possible outcomes of the draw. (Lots of different notations are used for the binomial coefficient; the first one is my preference.) The number of ways to draw exactly 2 orange marbles is
![\dbinom70\dbinom{10}2=1\cdot1\cdot45=45](https://tex.z-dn.net/?f=%5Cdbinom70%5Cdbinom%7B10%7D2%3D1%5Ccdot1%5Ccdot45%3D45)
That is, of the 3 red and 4 blue marbles - or the 7 non-orange marbles - we want 0; of the 10 oranges, we want 2.
So the probability of drawing exactly 2 orange marbles is
.
2) Now Jacob draws 3 marbles, for which there are
![\dbinom{17}3=680](https://tex.z-dn.net/?f=%5Cdbinom%7B17%7D3%3D680)
possible combinations. The event that Jacob draws at least 1 orange marble is complementary to the event that Jacob draws 0 orange marbles. So if we find the probabilty of drawing 0 oranges, then we subtract this from 1 to find the probability of drawing at least 1.
There are
![\dbinom73\dbinom{10}0=35\cdot1=35](https://tex.z-dn.net/?f=%5Cdbinom73%5Cdbinom%7B10%7D0%3D35%5Ccdot1%3D35)
possible ways of doing, so the probability of drawing 0 oranges is
, in turn making the probability of drawing at least 1,
.
3) This question is kinda ambiguous. It's not clear whether Jacob draws a marble with or without replacement.
If he does return a drawn marble to the bag, then for each draw, there is a
probability that he draws an orange marble, and a
probability of not. The draws are presumably independent of one another, so that the probability of drawing the first orange marble on the fourth attempt would be
![\dfrac7{17}\cdot\dfrac7{17}\cdot\dfrac7{17}\cdot\dfrac{10}{17}=\dfrac{3430}{83521}\approx0.041](https://tex.z-dn.net/?f=%5Cdfrac7%7B17%7D%5Ccdot%5Cdfrac7%7B17%7D%5Ccdot%5Cdfrac7%7B17%7D%5Ccdot%5Cdfrac%7B10%7D%7B17%7D%3D%5Cdfrac%7B3430%7D%7B83521%7D%5Capprox0.041)
If no replacements are made, then as non-orange marbles are drawn from the bag, the number of non-oranges and the total number of marbles both decrease by 1. Then the probability of drawing orange on the fourth draw would be
![\dfrac{\dbinom71\dbinom61\dbinom51\dbinom{10}1}{\dbinom{17}1\dbinom{16}1\dbinom{15}1\dbinom{14}1}=\dfrac5{136}\approx0.037](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cdbinom71%5Cdbinom61%5Cdbinom51%5Cdbinom%7B10%7D1%7D%7B%5Cdbinom%7B17%7D1%5Cdbinom%7B16%7D1%5Cdbinom%7B15%7D1%5Cdbinom%7B14%7D1%7D%3D%5Cdfrac5%7B136%7D%5Capprox0.037)
If you're supposed to round to the nearest hundredths place, your final answer should be acceptable either way.
4) We want the probability that a boat is made of wood given that it has chrome accents:
![P(\text{wood}\mid\text{chrome})=\dfrac{P(\text{wood AND chrome})}{P(\text{chrome})}=\dfrac{70\%}{75\%}\approx93\%](https://tex.z-dn.net/?f=P%28%5Ctext%7Bwood%7D%5Cmid%5Ctext%7Bchrome%7D%29%3D%5Cdfrac%7BP%28%5Ctext%7Bwood%20AND%20chrome%7D%29%7D%7BP%28%5Ctext%7Bchrome%7D%29%7D%3D%5Cdfrac%7B70%5C%25%7D%7B75%5C%25%7D%5Capprox93%5C%25)
5) We apply the inclusion/exclusion principle:
![P(\text{wood OR chrome})=P(\text{wood})+P(\text{chrome})-P(\text{wood AND chrome})](https://tex.z-dn.net/?f=P%28%5Ctext%7Bwood%20OR%20chrome%7D%29%3DP%28%5Ctext%7Bwood%7D%29%2BP%28%5Ctext%7Bchrome%7D%29-P%28%5Ctext%7Bwood%20AND%20chrome%7D%29)
![P(\text{wood OR chrome})=90\%+75\%-70\%=95\%](https://tex.z-dn.net/?f=P%28%5Ctext%7Bwood%20OR%20chrome%7D%29%3D90%5C%25%2B75%5C%25-70%5C%25%3D95%5C%25)
6) The events of a boat having chrome accents and a boat being made of wood are independent if and only if
![P(\text{wood AND chrome})=P(\text{wood})\cdot P(\text{chrome)}](https://tex.z-dn.net/?f=P%28%5Ctext%7Bwood%20AND%20chrome%7D%29%3DP%28%5Ctext%7Bwood%7D%29%5Ccdot%20P%28%5Ctext%7Bchrome%29%7D)
But
, so these two events are not independent.