We Can Form Them in Two Ways:

Question 1:

First of all, let us understand the concept. We have compound events which can be found on composition of two or more events.

We can form them in two ways: -

Union – Union of two events A and B is denoted by AUB and it is read as A or B and this means either A occurs or B occurs or both occur on any experiment.
Intersection – Intersection of two events A and B is denoted by AΠB. It is inverted U only in between. It is read as A and B and it means both A and B occurs on the same experiment.

And we have an important formula which is P(AUB) = P(A) +P(B) – P(AΠB) which can be seen in the following diagram.

Now let us move in solving the problems.

(a)We can find P(AUB) by simply using the above formula which is P(AUB) = P(A) +P(B) – P(AΠB)

So that becomes P(AUB) = 0.5+0.55-0.1 = 0.95

(b)P(A bar) = 1 – P(A). P(A bar) means probability of not happening of event A. So that becomes 1 – 0.5 = 0.5

(c)P(B bar) = 1 – P(B) = 1-0.55 = 0.45

(d)P(A intersection B bar) = P(A Alone) = 0.50 - 0.10 and P(B Alone) = 0.55 - 0.10 So P(A and (Not B)) = P(A Alone) = 0.40

(e)P(A intersection B) bar = 1- P(AΠB) = 1 -0.1 = 0.9

Question 2:

(a)Each time ball is orange (which occurs 7 of 19 possible times) 3 coins are tossed resulting in 8 possible cases HHH HHT HTH HTT THH THT TTH TTT. So that makes it 8*7 = 56

(b)The total number of elements in space will be 56 + for 2 tosses, it will be HH,TT,HT or TH. Therefore 56 + 4*12 = 56+48 = 104

Question 3:

(c)To have no heads, then you would have to have all tails. So there are 7 ways or we can explain in this way P(No Heads) = (7 0)(1/2)^0 (1/2)^7 = 1/128. SO elements would be (1/128)* 128 = 1 way

(d)Start or end (or both) with a head and have a total of exactly two heads. For this, we split it up into three tasks:

First one is heads, and last one isn't - We just need to put a single head among the remaining 5 spots, as the rest will be tails.

First one is heads and last one is heads - There is only one way to do this, first and last are heads and the rest are tails.

First one isn't heads, and last one is - This is identical to (1). So there are five ways.

Now we add all 3, we get 5+1+5 = 11

(e)Start and end with a head with a total of exactly three heads. For this, we know the first and last have a head, so we just put a head among the remaining 5 in the middle. So it is just 5.

Question 4:

A card is drawn at random from a regular playing card deck of 52 cards. Find the probability that a spade is drawn.

Solution: There are 13 spades in a deck of cards. So, the probability is Desired/total = 13/52 = ¼ = 0.25

Question 5:

All that is left in a packet of candy are 9 reds, 3 greens, and 2 blues.

(a)What is the probability that a random drawing yields a blue followed by a green assuming that the first candy drawn is put back into the packet?

Solution: P(blue then green) = P(blue) · P(green) =

(b)Are the events 'blue' and 'green' independent?

Solution: Because the first candy is replaced, the size of the sample space (14) does not change from the first drawing to the second so the events are independent.

Question 6:Employment data at a large company reveal that 54 % of the workers are married, that 38 % are college graduates and that 1/5 of the college graduates are married.
What is the probability that a randomly chosen worker is:
a) neither married nor a college graduate?

We have P(married) = 0.54, P(College graduate) = 0.34 and P( married and college graduate) = 0.38/5 = 0.076

P( neither married nor a college graduate ) = 1 – P( married or college graduate )

And P( married or a college graduate ) = [ P(married) + P(college graduate) – P( married andcollege graduate ) ]

= 0.54+0.38 – 0.076 = 0.844

Now P( neither married nor a college graduate ) = 1 – P( married or college graduate )

= 1 – 0.844 = 0.156 this is 15.6%

b) married but not a college graduate?

P( married but not a college graduate ) = P(married) – P( married and college graduate )

= 0.54 – 0.076 = 0.464 = 46.4%

c) married or a college graduate?

P( married or a college graduate ) = [ P(married) + P(college graduate) – P( married and college graduate ) ]

= 0.54+0.38 – 0.076 = 0.844

Question 7: Consider a sample space of three outcomes A,B,and C. Which of the following represent legitimate probability models?

A. P(A)=0.2, P(B)=0.2,P(C)=0.5 => => Not possible as cannot add up to 1. It is less than 1.
B. P(A)=-0.6, P(B)=0.7, P(C)=0.9 => Not possible as negative probability not possible.
C. P(A)=0.1, P(B)=0.9, P(C)=0 => It is legitimate as add up to 1
D. P(A)=0.4, P(B)=0, P(C)=0.6 => It is legitimate as add up to 1
E. P(A)=0.1, P(B)=0.9, P(C)=0.2. => Not possible as cannot add up to 1. It is more4 than 1.

Question 8: (Note that an Ace is considered a face card for this problem)
(a) P( black or a face card ) = P(Black) + P(Face card) –P(Black and Face)

We have 26 black cards so P(Black) = 26/52 and 16 face cards so P(Face card) = 16/52. There are 8 cards which are black and face cards so P(Black and Face) = 8/52.

So P( black or a face card ) = 26/52 + 16/52 – 8/52 = 34/52 = 17/26

(b) P( Queen and a 3) = Since 3 is not a queen so P( Queen and a 3) = 0

(c) P( black and a face card ) = There are 8 black cards which are Faces which are A,A,J,J,Q,Q,K,K. So probability is 8/52 = 2/13

(d) P( black and a Queen ) = There are two black queens so it is 2/52 = 1/26.

(e) P( face card or a number card ) = P(Face card) + P(Number card) –P(Face and Number). There are 16 face cards and 36 number cards and no cards in common so P( face card or a number card ) = 16/52 + 36/52 - 0 = 52/52 = 1

Question 9:

Real estate ads suggest that 60 % of homes for sale have garages,
29 % have swimming pools, and
26 % have both features.
What is the probability that a home for sale has
a) a pool or a garage?

P(Pool or Garage) = P(pool) + P(Garage) – P(Pool and Garage)

= 0.29 + 0.60 – 0.26 = 0.63

So 63% will have a pool or Garage

b) neither a pool nor a garage?

P(Neither a pool nor a garage) = 1 – P(Pool or a Garage)

= 1 – 0.63 = 0.37 =37%

37% of the homes have neither a pool nor a garage.

c) a pool but no garage?

P(Pool and no garage) = P(Pool) – P(Pool and Garage)

= 0.29 – 0.26 = 0.03

3% of the homes have a pool but no garage

Question 10 :

(a)A and B are Mutually Exclusive
∴ P(A∩B) = 0

(b) P(A or B) = P(A) + P(B) - P(A∩B)

= 0.55+0.35 -0 = 0.90

(d) P(not B) = 1- P(B) = 1- 0.35 = 0.65

(e) P(not(A or B)) = 1- P(A or B) = 1-0.90 = 0.10

(f) = P(A and (not B)) = P(A) - P(A∩B) = 0.45-0 =0.45

Question 11:There are five Oklahoma State Officials: Governor (G), Lieutenant Governer (L), Secretary of State (S), Attorney General (A), and Treasurer (T). Take all possible samples without replacement of size 3 that can be obtained from the population of five officials. (Note, there are 10 possible samples!)
(a) What is the probability that the governor and the treasurer are included in the sample

Solution: First, let's see how many possible samples would include the governor. To do this, we assume that the governor is part of the sample, which leaves the other two positions to be taken from the four remaining officials. 4!/(2! 2!) = (4*3)/(2*1) = .6. Six possible samples with the governor divided by ten total possible samples equals 60% chance he is included. Or required probability = (ways of having G & 2 others from 4) / 10
= 4C2/10
= .6

(a)What is the probability that the governor and the attorney general are included in the sample?

Solution: We have required probability = (ways of having Governor and Attorney General & 1 from 3 others) /10
= 3C1/10 = 0.3

Question 12: Two six-sided dice are rolled (one red and one green). Some possibilities are (Red=1,Green=5) or (Red=2,Green=2) etc.
(a) How many total possibilities are there?

Solution: First die = 6 possibilities
Second die = 6 possibilities
6*6 = 36 possibilities and does the order of red and green make a difference? In otherwords is (red=1,green=1) different than (green=1,red=1)?
if so, 6*6*2= 72 possibilities

(b) What is the probability that the sum on the two dice comes out to be 9?

Solution: Die 1 = 3,4,5,6 Die 2 =6,5,4,3
Die 1 = 6,5,4,3Die 2 = 3,4,5,6. So, 8 possibilities
8/36 = 1/4 < ANSWER

(b)What is the probability that the sum on the two dice comes out to be 12?

Solution:

Die 1: 6 Die 2: 6 = 1
1/36 < ANSWER

(c)What is the probability that the numbers on the two dice are equal?

Solution:Obviously there are 6 possibilities.
6/36 = 1/6 <ANSWER