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=(P(17,2))/(2!)=(P^(17)_2)/(2!)=(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
`(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
`(35)/(680)`, in turn making the probability of drawing at least 1,
`1-(35)/(680)=(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
`(10)/(17)` probability that he draws an orange marble, and a
`\frac7{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

`\frac7{17}\cdot\frac7{17}\cdot\frac7{17}\cdot(10)/(17)=(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

`\frac{\dbinom71\dbinom61\dbinom51\dbinom{10}1}{\dbinom{17}1\dbinom{16}1\dbinom{15}1\dbinom{14}1}=\frac5{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})=\frac{P(\text{wood AND chrome})}{P(\text{chrome})}=(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\%\\eq70\%`, so these two events are not independent.