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1) A sample of 3 items is selected at random from a box containing 20 items of which 4 are defective. Find the expected number of defective items:

Since there is a sample of 3 and a total of 20 items, i can use a hyper-geometric distribution:

N: 20

n: 3

x: number of success in sample n, ?

k: number of success in population N, ?

E[x]=$fracn times kN$

Originally I had this as a binomial with P: (1/5) 4 success in 20 trials...

2) Suppose a player is dealt 5 cards. Find the probability that a player is dealt a pair: This could be done with a hyper geometric I believe:

N: 52

n: 5

x: 2

k: 3?

3) A box contains 5 red and 5 blue marbles. 2 are withdrawn randomly. If they are the same color you win 1.10 if they are different you lose 1.00. Calculate the expected value you win and the variance:

Each time a ball is drawn the event is independent with a probability of 1/2 of success and 2 trials:

E[x]= 2(1/2)=1

Var[x]=np(1-p)=2(1/2)(1/2)=.50

4) Suppose the average number of cars abandoned weekly on a certain highway is 2.2. Approximate the probability that there will be a) no abandoned b) at least 2

a) Poisson with $mu=2.2$ and $x=0$:

$frace^-2.2 times 2.2^00!$

b) 1-(P(0)+P(1)+P(2)+P(3)) Im not going to calculate its a pain but the reasoning is correct?

A little help on some parts, where you have shown some of your work.

$1.$ Hypergeometric is correct. Expected value 0.6, as you say.

$2.$ You show nothing to support your 'hypergeometric' guess.

$3.$ Suggest you use a tree diagram. The probability both balls are red is
$P(R_1 cap R_2) = P(R_1)P(R_2|R_1) = (5/10)(4/9)= 2/9.$
Then find the probability both are green. Add to get the probability
they are both of the same color. Finally, figure the distribution of
your winnings.

$4$(b). $X sim mathsfPois(lambda=2.2)$ and you seek
$$P(X ge 2) = 1 - P(X le 1) = 1 - P(X = 0) - P(X = 1).$$
You already have $P(X = 0)$ from part (a), so it's not 'such a pain'
as you thought.