Question

I don’t get any of these :/// please help

Answer 1

1) From the total of 17 marbles, Jacob takes 2 out at random. There are

$\dbinom{17}2=C(17,2)=C^{17}_2=\dfrac{P(17,2)}{2!}=\dfrac{P^{17}_2}{2!}=\dfrac{17!}{2!(17-2)!}=136$

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

$\dbinom70\dbinom{10}2=1\cdot1\cdot45=45$

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 $\dfrac{45}{136}\approx0.33$.

2) Now Jacob draws 3 marbles, for which there are

$\dbinom{17}3=680$

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

$\dbinom73\dbinom{10}0=35\cdot1=35$

possible ways of doing, so the probability of drawing 0 oranges is $\dfrac{35}{680}$, in turn making the probability of drawing at least 1, $1-\dfrac{35}{680}=\dfrac{645}{680}\approx0.95$.

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 $\dfrac{10}{17}$ probability that he draws an orange marble, and a $\dfrac7{17}$ 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

$\dfrac7{17}\cdot\dfrac7{17}\cdot\dfrac7{17}\cdot\dfrac{10}{17}=\dfrac{3430}{83521}\approx0.041$

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

$\dfrac{\dbinom71\dbinom61\dbinom51\dbinom{10}1}{\dbinom{17}1\dbinom{16}1\dbinom{15}1\dbinom{14}1}=\dfrac5{136}\approx0.037$

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:

$P(\text{wood}\mid\text{chrome})=\dfrac{P(\text{wood AND chrome})}{P(\text{chrome})}=\dfrac{70\%}{75\%}\approx93\%$

5) We apply the inclusion/exclusion principle:

$P(\text{wood OR chrome})=P(\text{wood})+P(\text{chrome})-P(\text{wood AND chrome})$

$P(\text{wood OR chrome})=90\%+75\%-70\%=95\%$

6) The events of a boat having chrome accents and a boat being made of wood are independent if and only if

$P(\text{wood AND chrome})=P(\text{wood})\cdot P(\text{chrome)}$

But $90\%\cdot75\%=67.5\%\neq70\%$, so these two events are not independent.