A standard deck of cards has 52 cards. The cards have one of two colors: 26 cards in the deck are red and 26 are black. The cards have one of four denominations: 13 cards are hearts (red), 13 cards are diamonds (red), 13 cards are clubs (black), and 13 cards are spades (black).

a. One card is selected at random and the denomination is recorded. What is the sample space S for the set of possible outcomes?
b. Two cards are selected at random and the color is recorded. What is the sample space S for the set of possible outcomes?
c. Two cards are selected at random and the denomination is recorded. The event H is defined as the event that the first card is hearts. What defines event H?
d. Two cards are selected at random and the denomination is recorded. The event D is defined as the event that the first card is diamonds and the second card is red. What defines event DC?
e. Two cards are selected at random. Event C is defined as the event that the first card is clubs, event R as the event that the first card is red, and event B as the event that the second card is black. Which events are disjoint?

Given:

The cards have one of four denominations:

13 cards are hearts (red),

13 cards are diamonds (red),

13 cards are clubs (black), and

13 cards are spades (black).

a) One card is selected at random and the denomination is recorded. What is the sample space S for the set of possible outcomes?

Sample sample S = {heart, diamond, spade, club} as we only have four types of denominations. It means we can get one of the denominations at a a time, that is why the we have the sample space S.

b) Two cards are selected at random and the color is recorded. What is the sample space S for the set of possible outcomes?

Here, we are talking about colors, and we have two types of color black and red, so the sample space will be

As we are selecting two cards at random, so each element in the sample space will be the combination of two colors, and the set of possible outcomes are 4 (2^2 = 4) which is

S = {(black, black), (black, red), (red, black),(red, red)}

c)Two cards are selected at random and the denomination is recorded. The event H is defined as the event that the first card is hearts. What defines event H?

Here, we are talking about denominations, and we have four types of denominations, so the sample space H will be

As given, the first card is heart,

H = {(heart, heart), (heart, diamond), (heart, spade), (heart, club)}

we can get either of the combination from H provided in all combination first card is heart, and other card could be any of the four denomination. .

d)Two cards are selected at random and the denomination is recorded. The event D is defined as the event that the first card is diamonds and the second card is red. What defines event DC?

Here, we are talking about denominations and colors, and we have four types of denominations and two colors,

As given first card is diamond it second card will be of red color and we know that heart and diamond in the deck of cards are of red color, so the sample space D will be

D = {(Diamond, heart), (diamond, diamond)}

It means we have only two options.

e)Two cards are selected at random. Event C is defined as the event that the first card is clubs, event R as the event that the first card is red, and event B as the event that the second card is black. Which events are disjoint?

Disjoint events in probability means that the events have no outcomes in common.

For event C, first card is club, so the outcome is

C = {(club, heart), (club, diamond), (club, spade), (club, club)}

For event R, first card is red, so the outcome is

R = {(red, black),(red, red)}

For event B, second card is black, so the outcome is

B = {(red, black),(black, black)}

Here we can see events R and B have one outcome as common which is {red, black} and C has nothing in common among all.

So the event C is disjoint.  

(1+1+1)!=6

2+2+2=6

(3×(3/3)!=6

(4-(4/4))!=6

5+(5/5)=6

6*(6/6)=6

7-(7/7)=6

((∛8)+(8/8))!=6

(√9×(9/9))!=6

Step-by-step explanation:

Since no restrictions were given on what operations to use the factorial operation is just n!=n×(n-1)×(n-2)...1:

These are the arithmetic steps:

(1+1+1)!=3!=3×2=6

2+2+2=6

(3×(3/3))!=3!=3×2=6

(4-(4/4))!=(4-1)!=3!=3×2=6

5+(5/5)=5+1=6

6*(6/6)=6*1=6

7-(7/7)=7-1=6

((∛8)+(8/8))!=(2+1)!=3!=3×2

(√9×(9/9))!=(3×1)!=3×2=6