Question

1. In a list of 15 households, 9 own homes and 6 do not own homes. Five households are randomly selected from these 15 households. Find the probability that the number of households in these 5 own homes is exactly 3.

Round your answer to four decimal places

P (exactly 3)=

2. A really bad carton of 18 eggs contains 9 spoiled eggs. An unsuspecting chef picks 6 eggs at random for his " Mega-Omelet Surprise." Find the probability that the number of spoiled eggs among the 6 selected is

(A) exactly 6. Round your answer to four decimal places

(B) 2 or fewer. Round your answer to four decimal places

(C) more than 1. Round your answer to four decimal places.

3. Six jurors are to be selected from a pool of 20. potential candidates to hear a civil case involving a lawsuit between two families. unknown to the judge or any of the attorneys, 4 of the 20 prospective jurors are potentially prejudiced by being acquainted with one or more of the litigants. they will not disclose this during the jury selection process. if 6 jurors are selected at random from this group of 20, find the probability that the number of potentially prejudiced jurors among the 6 selected jurors is exactly 1

Round your answer to four decimal places

5. you get a fruit basket containing 10 apples. let X be the number of apples left in the basket. Determine all possible values for the random varible X?

6. you grab two books at random off a shelf . let X be the number of those two books that you have previously read . determine all possible values for the random variable x?

7. Classify each variable as discrete or continuous

(a) the weight of a bunch of bananas
(b) the number of bananas in a bunch
(c) the time it takes to run a mile
(d) the number of push-ups a person can do

8. A random variable is a variable whose value is determined by the?

9.A discrete random variable is a random variable?

Answer 1

#1.
5 households from 15 can be selected in 15C5.
3 households which own homes can be selected in 9C3 and remaining 2 which do not can be selected in 6C2

Hence required probability,
P(Exactly 3) = 9C3*6C2/15C5 = 0.5594

#2.
total 18, 9 - spoiled and 9 - good

A)
9C6 indicates all selected are spoiled
P(X = 6) = 9C6/18C6 = 0.0045

B)
P(X <= 2) = (9C6*9C0 + 9C5*9C1 + 9C4*9C2)/18C6
P(X <= 2) = 0.3100

C)
P(X > 1) = 1 - P(X <= 1)
P(X <= 1) = (9C6*9C0 + 9C5*9C1)/18C6 = 0.0656
P(X > 1) = 1 - 0.0656 = 0.9344