Misc
How many 6 digit numbers can be formed using 0-9 (Repetition not allowed)
9*9*8*7*6*5 = 136080
One card is drawn from a standard deck.
Given that the card is a spade, what is the probability that it is a face card (J, Q, K)?
There are 13 spades. Face cards in spades: J♠, Q♠, K♠ → 3 cards.
P(face∣spade)=3/13
There are 8 runners in a race. In how many different ways can the top 3 runners line up for gold, silver, and bronze medals?
P(8,3)=8! / (8−3)! = 5! * 8! =336
a) No gym membership
b) Has a swimming membership
a) No gym total = 150+150=300.
P(Non Gym) = 300/600 = 1/2
b) Swimming total = 180 + 150 = 330
P(Swimming) = 330/600 = 11/22 = .55
A jar contains 5 green and 7 yellow balls. Two balls are drawn with replacement. What is the probability that both are green?
With replacement:
P=5/12⋅5/12=25/144≈0.1736
Two cards are drawn from a standard 52-card deck without replacement.
a) What is P(2nd card is a heart∣1st card is a heart)P(\text{2nd card is a heart} \mid \text{1st card is a heart})P(2nd card is a heart∣1st card is a heart)?
b) What is P(1st card is a heart∣2nd card is a heart)P(\text{1st card is a heart} \mid \text{2nd card is a heart})P(1st card is a heart∣2nd card is a heart)?
There are 13 hearts in a deck.
a) Given the first card is a heart, there are 12 hearts left out of 51 cards:
P(2nd heart∣1st heart)=12/51=4/17≈0.23529.
b) By symmetry (or compute similarly): given the second is a heart, probability the first was a heart is also
P(1st heart∣2nd heart)=12/51=4/17≈0.23529.
A restaurant lets you build your own lunch special by choosing:
3 types of sandwiches (turkey, ham, veggie),
2 types of sides (chips, salad),
4 types of drinks (soda, water, tea, lemonade).
If you choose one sandwich, one side, and one drink, how many different lunch specials can be made?
3×2×4=24
c) Gym and no Swimming
d) Gym or Swimming
c) Gym and No Swim = 120
P(gym and no swim) = 120/600 = 1/5 = .2
d) Use P(A U B) = P(A) + P(B) - P(A N B)
P(gym) = 180 + 120 / 600 = 300/600 = 1/2
P(swim) = 330/600
P(gym and swim) = 180/600
P(gym or swim)=1/2 + 330/600 - 180/600 = 300+330-180 / 600 = 450/600 = 3/4 = .75
In a class of 20 students, 12 have a dog, 8 have a cat, and 5 have both a dog and a cat.
Question: If a student is chosen at random, what is the probability that the student has a dog given that they have a cat?
Given:
Students with dog = 12
Students with cat = 8
Students with both = 5
We want:
P(dog | cat)=P(dog AND cat)/P(cat)
P(dog AND cat)=520=0.25
P(cat)=820=0.4
P(dog | cat)=0.25/0.4=0.625
A coin is flipped twice.
Let A= “first flip is heads,” and B= “second flip is heads.”
Find P(A∣B).
Sample space = {HH, HT, TH, TT}.
B={HH,TH}, ∣B∣=2.
A∩B={HH},∣A∩B∣=1
So
P(A∣B)=∣A∩B∣ / ∣B∣=1/2
A student council has 12 members. The school principal wants to select 4 students to represent the council at a leadership conference.
(4 C 12)= 4! (12−4)! 12!=4!⋅8!*12! = 495
e) P(gym ∣ swim)
f) P(no gym ∣ swim)
e)
P(no gym ∣ swim) = P(gym and swim) / P(swim) = (180/600) / (330/600) = 180/330 = 6/11
f)
No gym and swim = 150
P(no gym | swim) = (150/600) / (330/600) = 150/330 = 5/11
A bag contains 5 red, 3 blue, and 2 green marbles. One marble is drawn at random.
Question: What is the probability that the marble is red or green?
P(red or green)= 7/10
A card is drawn from a standard 52-card deck.
Let E= “card is red,” and F= “card is a king.”
Find P(E∣F)
There are 4 kings total, 2 are red (♥, ♦).
So
P(E∣F)=2/4=1/2
A bookshelf has 7 different books. In how many ways can you arrange 3 of them on the shelf?
7P3 = 7! / (7−3)!= 4! * 7! = 210
Are gym and swim independent?
Are P(gym and swim) = 180/600 = .3
P(gym)P(swim) = .5*.55 = .275
They are not equal, so the events are not independent
A standard deck of 52 playing cards is used. One card is drawn at random.
Question: What is the probability that the card is either a heart or a face card (Jack, Queen, King)?
P(A or B)=P(A)+P(B)−P(A and B)
P(heart or face card)=22/52=11/26
A bag has 3 blue and 2 red marbles (5 total). One marble is drawn at random.
Let G= “marble is blue,” and H= “marble is not red.”
Find P(G∣H)
H={not red}={blue}, so 3 outcomes.
All of those are blue.
So
P(G∣H)=3/3=1.
A teacher has a set of 15 books and wants to choose 5 of them to bring on a trip.
How many different groups of books can the teacher choose?
(15 C 5) = 15! / 5!(15−5)!= 15! / 5!⋅10! = 3003
There are 12 teachers and 18 students in a math club. The club will send 6 representatives to a regional competition.
a) How many ways are there to select a group of 6 people?
(a) Total ways to select 6 people
We are selecting 6 people out of 30 without regard to order. This is a combination problem:
Total ways=(30 6)
(30 C 6)=30⋅29⋅28⋅27⋅26⋅25.../ 6⋅5⋅4⋅3⋅2⋅1 = 593,775
✅ Answer (a): 593,775 ways