Question

More helppp

Answer 1

1. So the total number of outcomes is 20.
a. P(2 or 5): There are 2 favorable outcomes {2, 5}... so P = 2/20 = 1/10

b. Multiples of 2: {2, 4, 6, 8, 10, 12, 14, 16 18, 20} these are the 10 numbers
Multiples of 5: {5, 10, 15, 20} these are the 4 numbers
(SINCE 12 AND 20 ARE IN BOTH LISTS, WE ONLY COUNT THEM ONCE)
Unique outcomes: {2, 4, 5, 6, 8, 10, 12, 14, 15, 16, 18, 20} these are the 12 numbers

So your final result is P = 12/20 = 3/5


2. So a standard card deck has 13 cards of each suit

a. P(heart or club): 13 hearts + 13 clubs = 26 cards. P = 26/52 = 1/2
b. P(heart and club): A single card cannot be two suits at once. P = 0/52 = 0
c. P(red or heart): All hearts are already red. So this is just the total number of red cards, which is 26. P = 26/52 = 1/2
d. P(jack or heart): 13 hearts + 3 other jacks.P = 16/52 = 4/13
e. P(red or black): Every card is either red or black. 52/52 = 1
f. P(red queen or black jack): 2 red queens + 2 black jacks = 4 cards. 4/52 = 1/13
g. P(red or ten): 26 red cards + 2 black tens. P = 28/52 = 7/13
h. P(spades or a face card): 13 spades + 9 face cards from the other suits. P = 22/52 = 11/26




3. So the total number of cards is 7

a. P(J or K): 2 favorable outcomes ({J, K}). So P would = 2/7

b. P(L or M or N): 3 favorable outcomes ({L, M, N}). So P would = 3/7

c. P(O or a vowel): In this set, "O" is the only vowel. So P would = 1/7

Answer 2

and 4 and 5?

Answer 3

For #4: To find the number of students who did both (community service AND honor roll), we use the Inclusion-Exclusion Principle formula:

First we're gonna identify the given values:

Total students (N): 500
Students with award (A U B): 180
Community Service (A): 125
Honor Roll (B): 75

Now we're going to solve for the overlap:

180 = 125 + 75 - Both

180 = 200 - Both

Both = 200 - 180 = 20

Now we're going to calculate the probability:

P(Both) = 20/500 = 2/50 = 0.04 (or 4%)




For #5:

1. True: Every set is a subset of itself.

2. False: While Rui and Jurgen are in A, Bale is not in set A. Therefore, C is not a subset of A.

3. False: Will is in sets A and B, but he is not listed in set C.



For #6:

1. Since every person in $B$ is also in $A$, the intersection is just set B {Will, Karmen, Rui}

2. Combine all names from both lists, removing duplicates (Rui appears in both):
{Will, Karmen, Rui, Bale, Jurgen}

3. Assuming set A (or the combination of all named students) is the universal set, we list the names in A that are not in C: {Will, Julio, Todd, Karmen, Kiki}

4. First, find B ∩ C: Only {Rui} is in both.The complement is everything in the group except Rui. {Will, Julio, Todd, Karmen, Jurgen, Kiki, Bale}

Answer 4

@randogirl123

Answer 5

u lost me

Answer 6

@randogirl123 hope that helped

Answer 7

I do astronomy and I got so lost n math