Basic Probability
In a box, there are 5 blue marbles and 3 green marbles. You draw three marbles from the box with replacement. What is the probability that exactly two of the marbles you have drawn are green.
There are 3 ways to draw two green and 1 blue marble: GGB, GBG, and BGG.
Each way has probability (3/8)(3/8)(5/8)=45/512
Therefore, the probability of drawing two green and one blue marble is
3xx(3/8)(3/8)(5/8)=135/512
Josephine is making a playlist to listen to while she does her gardening. She has 350 songs in her library, but only wants 20 songs on her playlist. She is using a randomizer to select and order the 20 songs that will go on her playlist from her total of 350 songs.
Write (but do not evaluate) an expression to represent the number of distinct playlists that can be generated in this manner.
Since putting songs in different orders will necessarily create different playlists, we must use a permutation rather than a combination. In other words, the order is important in this situation.
The number of distinct playlists that can be generated is enormous, but can be represented by the expression,
(350!)/((350-20)!)
In a society where cats are the apex animal life form, 50% of cat families keep humans as pets, 30% keep mice as pets, and 40% of cat families do not keep humans or cats as pets.
Given that a randomly-selected cat family keeps human pets, find the probability that they also keep mice as pets.
First, note that
(HcapM)=0.2
Then,
P(M|H)=(P(HcapM))/(P(H))=0.2/0.5=2/5
What is the coefficient on the 4th term in the expansion of (x+y)^6 when the terms of the expansion are listed from highest power of x to lowest power of x?
(x+y)^6=x^6 + 5x^5y + 15x^4y^2 +20x^3y^3 + 15x^2y^4 + 5xy^5 + y^6
The fourth term in this expansion is
20x^3y^3
The coefficient of which can be found by,
((6),(3))={6!}/{3!xx3!}=20
In an experiment, a student rolled a 6-sided die 200 times and recorded a mean score of 4.4.
Determine whether or not this die is fair. Offer sound mathematical justification for your answer.
The die is most likely not fair. In a fair die, the probability of rolling each number is 1/6 . Therefore, the expected value of the distribution of scores should be
sum_(i=1)^6 x_ip_i=(1+2+3+4+5+6)/6=3.5
From a class of 18 students, 3 are to be chosen randomly to represent their grade level in a student vs. teacher basketball match. Of these 18 students, 5 are on the school basketball team.
Find the probability that of the 3 randomly-chosen students from this class, at least two of them are on the school basketball team.
First find the probability that there exactly two basketball team members are chosen:
P(2BB)=((3),(2))((5 xx4xx 13)/(18xx17xx16))=65/408
Then, find the probability that all three chosen are basketball players:
P(3BB)=(5xx4xx3)/(18xx17xx16)=5/408
Finally,
P(2BB)+P(3BB)=70/408=35/204
A student rolls one 6-sided die and one 8-sided die, and sums the results of the two dice.
Calculate the probability that the sum is either 10 or 4.
Express your answer as a reduced fraction.
5/48 + 3/48 = 8/48 = 1/6
Laurie has 5 pairs of red socks and one pair of pink socks. It's laundry day, and she's hanging up all her socks to dry.
In how many visually distinct ways can she hang up her red and pink socks?
Given 12 total socks, there are ((12),(2)) ways to choose two of them to be pink.
Therefore, there are ((12),(2))=(12!)/(2!xx(12-2)!)= 66 visually distinct ways in which Laurie can hang her socks.
Let P(A)=2/5 , P(B)=1/3 , and P(AcupB)=1/2 .
Determine (with reasoning) whether or not A and B are independent events.
Recall that,
P(AcupB)=P(A)+P(B)-P(AcapB)
In this case,
1/2=2/5+1/3-P(AcupB)
This results in,
P(AcapB)=7/30
Then, since
P(A)xxP(B)=2/15 !=7/30
we have that A and B are not independent events.
Let A and B be two mutually exclusive events. Further let it be that A' and B' are also mutually exclusive.
Find P(AcupB)
Since A and B are mutually exclusive, we have,
A capB=O/
Then, since A' and B' are mutually exclusive, we have,
A'capB'=O/
Thus A and B partition the entire universe. That is, there is no element in the universe that is not in A union B.
Therefore,
P(AcapB)=1
Let S be the set of all six-digit integers made up only of the digits 1, 2, 3, 4, 5, and 6, and such that each of these digits is used only once per integer.
If the elements of S are listed in order from least to greatest, find the 363rd element in the list.
First, consider all elements of S such that the first digit is 1. Then, since there are five digits following the 1, there are 5! different ways to arrange them. Therefore, there are 5! =120 different elements of S with 1 as a first digit.
Likewise, there are 120 different elements with first digit 2, and 120 different elements with first digit 3. After accounting for all elements with first digit either 1, 2, or 3, there will be 360 elements.
Then the first element with first digit 4 is 412356.
This is the 361st element in the list.
The 362nd element would be 412365, and finally, the 363rd element would be 412536.
In the grade 9 cohort at a certain school, 54% of the student population are girls and the rest are boys. Of the girls, 45% are Cantonese speakers, and of the boys, 60% are Cantonese speakers.
If a student is chosen at random, calculate the probability that they are a boy, given that they are a Cantonese speaker. Round your answer to 3 significant figures.
P(GnnC)=0.540 xx0.450 = 0.243
P(BnnC)=0.46 xx0.06 = 0.276
P(C) = P(G nnC)+P(BnnC) = 0.243 + 0.276 = 0.519
P(B|C) = {P(BnnC)}/{P(C)} = 0.276/0.519~~0.532
There is a 90% chance that on any given day, the school cafeteria serves some dish containing seafood.
Calculate the probability that in the span of Monday through Friday, the school cafeteria serves a seafood dish no more than 3 times. Express your answer as a reduced fraction.
X~B(5, 9/10)
Then, P(X≤3)=1-[P(X=4)+P(X=5)]
P(X≤3)=1 - [((5),(4))(9/10)^4(1/10)+((5),(5))(9/10)^5]
=1-(32805/100000+59049/100000)
=4073/50000
Let A and B be two mutually exclusive events. Further let it be that A' and B' are also mutually exclusive.
Prove that it is impossible for both A and B to be empty.
(Hint: Assume that the universe is non-empty)
Since A and B are mutually exclusive, their intersection is the empty set.
Likewise, since A' and B' are mutually exclusive, their intersection is the empty set.
Assume instead that both A and B are empty. Thus, A' consists of the entire universe set. Likewise B' consists of the entire universe set. Let x be an element of the universe set. Then x is an element of A' and x is an element of B'. But this is a contradiction, since the intersection of A' and B' is empty.
Therefore, it is impossible that both A and B are empty.
Let A and B be independent events.
Prove that A' and B' are also independent events.
We need to show that
P(A'capB')=P(A')xxP(B')
By DeMorgan's law,
P(A'capB')=P(AcupB)'=1-P(AcupB)
Then, since A and B are independent,
=1-[P(A)+P(B)-P(A)P(B)]
=1-[(1-P(A'))+(1-P(B'))-(1-P(A'))(1-P(B')]
=1-1+P(A')-1+P(B')+1-P(B')-P(A')+P(A')xxP(B')
=P(A')xxP(B')
What is the coefficient on the x^2 term in the expansion of (3x+1/x)^8 ?
Need
((8),(k))(3x)^k(1/x)^{8-k}
such that
k-(8-k)=2 rArrk=5
Then,
((8),(5))(3x)^5(1/x)^{8-5}=13608x^2
Thus the coefficient is 13608.