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Problems involving sets, functions, and probability.
Published on February 27, 2025
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The solution finds the symmetric difference between two sets derived from inverse functions, demonstrating the application of set theory and function operations. It also proves a set identity involving unions and intersections of sets, and calculates derivatives of composite functions using the chain rule. Finally, it calculates probabilities of events involving coin tosses and dice rolls, and determines the probability of having 53 Sundays or Mondays in a leap year.
Question
Answer
Here are the step-by-step solutions to the math problems:
1. Consider two functions
First, we find the set
Taking the cube root of all parts, we get:
So,
Next, we find the set
Adding 3 to all parts, we get:
Multiplying all parts by 2, we get:
So,
Let
Therefore, the symmetric difference is:
Answer: The symmetric difference is
2. Consider the sets
First, we calculate
Then, we calculate
Next, we calculate
And
Finally, we calculate
Since
Answer: We have shown that both sides are equal to
3. Consider two functions
First, we find the composite function
Now, we differentiate
Using the chain rule, we have:
Answer:
4. A coin is tossed and a dice is rolled. Write down the sample space. What is the probability of:
i) getting odd number in the dice and head in the coin?
ii) getting prime number in the dice and tail in the coin?
i) getting odd number in the dice and head in the coin?
ii) getting prime number in the dice and tail in the coin?
The sample space
The total number of outcomes is
i) Event A: getting an odd number in the dice and head in the coin.
Odd numbers on the dice are
Favorable outcomes for A are
The probability of event A is
Odd numbers on the dice are
Favorable outcomes for A are
The probability of event A is
ii) Event B: getting a prime number in the dice and tail in the coin.
Prime numbers on the dice are
Favorable outcomes for B are
The probability of event B is
Prime numbers on the dice are
Favorable outcomes for B are
The probability of event B is
Answer:
Sample space:
i) Probability of getting odd number in the dice and head in the coin is
ii) Probability of getting prime number in the dice and tail in the coin is
Sample space:
i) Probability of getting odd number in the dice and head in the coin is
ii) Probability of getting prime number in the dice and tail in the coin is
5. Consider two functions
First, we find
Next, we find
Now, we find
Finally, we find
Answer:
6. Consider two sets
The power set of A,
The power set of B,
The symmetric difference
Answer:
7. Assume a box contains 7 red balls, 13 green balls, and 17 white balls. What is the probability of picking up (for both cases) without replacement:
i) a green ball followed by a white ball, and then another green ball?
ii) a green ball first, then a white ball, and finally a red ball?
i) a green ball followed by a white ball, and then another green ball?
ii) a green ball first, then a white ball, and finally a red ball?
Total number of balls =
i) Probability of picking a green ball, then a white ball, then a green ball.
P(1st ball is green) =
After picking one green ball, there are 36 balls left, with 12 green and 17 white.
P(2nd ball is white | 1st ball is green) =
After picking one green and one white ball, there are 35 balls left, with 12 green and 16 white and 7 red.
P(3rd ball is green | 1st ball is green, 2nd ball is white) =
The probability of picking a green ball, then a white ball, then a green ball is:
P(1st ball is green) =
After picking one green ball, there are 36 balls left, with 12 green and 17 white.
P(2nd ball is white | 1st ball is green) =
After picking one green and one white ball, there are 35 balls left, with 12 green and 16 white and 7 red.
P(3rd ball is green | 1st ball is green, 2nd ball is white) =
The probability of picking a green ball, then a white ball, then a green ball is:
ii) Probability of picking a green ball, then a white ball, then a red ball.
P(1st ball is green) =
P(2nd ball is white | 1st ball is green) =
After picking one green and one white ball, there are 35 balls left, with 12 green, 16 white and 7 red.
P(3rd ball is red | 1st ball is green, 2nd ball is white) =
The probability of picking a green ball, then a white ball, then a red ball is:
P(1st ball is green) =
P(2nd ball is white | 1st ball is green) =
After picking one green and one white ball, there are 35 balls left, with 12 green, 16 white and 7 red.
P(3rd ball is red | 1st ball is green, 2nd ball is white) =
The probability of picking a green ball, then a white ball, then a red ball is:
Answer:
i) The probability of picking a green ball followed by a white ball, and then another green ball is
ii) The probability of picking a green ball first, then a white ball, and finally a red ball is
i) The probability of picking a green ball followed by a white ball, and then another green ball is
ii) The probability of picking a green ball first, then a white ball, and finally a red ball is
8. What is the probability of having 53 Sunday or 53 Monday in a leap year?
A leap year has 366 days, which is 52 weeks and 2 extra days (
The possible pairs of extra days are: (Sunday, Monday), (Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday), (Saturday, Sunday). There are 7 equally likely possibilities.
The possible pairs of extra days are: (Sunday, Monday), (Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday), (Saturday, Sunday). There are 7 equally likely possibilities.
For having 53 Sundays, one of the extra days must be Sunday. This occurs in the pairs (Saturday, Sunday) and (Sunday, Monday).
For having 53 Mondays, one of the extra days must be Monday. This occurs in the pairs (Sunday, Monday) and (Monday, Tuesday).
For having 53 Sunday or 53 Monday, we need to count the cases where we have 53 Sundays or 53 Mondays or both. The favorable pairs are (Saturday, Sunday), (Sunday, Monday), (Monday, Tuesday). There are 3 favorable outcomes.
For having 53 Mondays, one of the extra days must be Monday. This occurs in the pairs (Sunday, Monday) and (Monday, Tuesday).
For having 53 Sunday or 53 Monday, we need to count the cases where we have 53 Sundays or 53 Mondays or both. The favorable pairs are (Saturday, Sunday), (Sunday, Monday), (Monday, Tuesday). There are 3 favorable outcomes.
The probability of having 53 Sunday or 53 Monday in a leap year is
Answer: The probability is
9. Consider two sets
A function from X to Y is a relation where each element in X is mapped to exactly one element in Y.
Relation 1 (Function
Relation 1 (Function
Relation 2 (Function
Both
Answer:
Two relations which are functions from X to Y are:
Two relations which are functions from X to Y are:
10. Write down two finite and two infinite sets in the tabular and set builder form.
Finite Sets:
Set 1:
Tabular form:
Set builder form:
Set 1:
Tabular form:
Set builder form:
Set 2:
Tabular form:
Set builder form:
Tabular form:
Set builder form:
Infinite Sets:
Set 1:
Tabular form:
Set builder form:
Set 1:
Tabular form:
Set builder form:
Set 2:
Tabular form:
Set builder form:
Tabular form:
Set builder form:
Answer:
Finite Sets:
Set 1: Tabular form:
Set 2: Tabular form:
Infinite Sets:
Set 1: Tabular form:
Set 2: Tabular form:
Finite Sets:
Set 1: Tabular form:
Set 2: Tabular form:
Infinite Sets:
Set 1: Tabular form:
Set 2: Tabular form: