We enumerate the crayons from 1 to 15. Let Xi indicate when the i^th crayon is among the 5 crayons selected.

So,
E(Xi) = Pr {Probability that at least one type i crayon is in set of 5}
E(Xi) = 1 – Pr {no type i crayons in set of 5}
E(Xi) = 1 – (14⁵ /15)

Therefore, the expected # of crayons is:

= ∑¹⁵ E(Xi)
= 15[1-(14⁵/15))]
= 4.38

#Define our array of target colors
fav_color_filter = ['blue', 'yellow']
#To select rows whose column value is in an iterable array, which we defined as fav_color_filter, we can use isin
df = df.loc[df['favorite_color'].isin(fav_color_filter)]
#next, we need to filter on scores above 90. here we can use loc on our dataframe:
df = df.loc[(df['grade'] >= 90)]
#preview the dataframe
df.head()

SELECT
# (3) Pull all records from our subquery below
* FROM
# (1) Here, we can create a subquery to stack rank the landfills by weight
# for each given state
(SELECT
    t.landfillID,
    t.weight,
    t.state,
    # (2) Using rank, we assign an integer ranking to each row,
    # allowing us to see which landfills have the most garbage
    # in them (and second most, etc) in each state
    RANK() OVER(PARTITION BY state ORDER BY t.weight DESC) as rank_weight
FROM landfill_weights t) stg
# (4) When pulling from our sub query, we simply filter where the rank_weight = 2
# to grab landfills with the second highest amount of garbage, per the question prompt
WHERE stg.rank_weight = 2