A. Identify the misconceptions, or problems in the provided solutions if any and B. Provide at two(2) remedies to the problems or/and misconceptions identified. b. Suppose we choose an integer from the set of the first ten 10 positive integers. The sample space of this experiment is the set S = 1, 2, 3, ..., 10 Each of the sets A = {2, 4, 6, 8, 10}, B = {1, 3, 5, 7, 9}, C = {3, 6, 9}, is an event. A is the event of choosing an even integer, B is the event of choosing an odd integer while C is the event of choosing an integer divisible by 3. The complement of C is is given by C 0 = {1, 2, 4, 5, 7, 8, 10}. The union of B and C is B ∪ C = {1, 3, 3, 5, 6, 7, 9, 9} and the number of elements in set B ∪ C is n(B ∪ C) = 8. The intersection of B and C is B ∩ C = {3, 3, 9, 9} and the number of elements in set B ∩ C is n(B ∪ C) = 4. Notice that (i) C ∪ C 0 = S (ii) C ∩ C 0 = A ∪ B The events B and C are mutually exclusive. b. We want to draw a diagram to show the vector below and then write it in magnitude-direction form. 3 −2 The magnitude of the vector is (hypotenuse)2 = (opposite)2 + (adjacent)2 h 2 = −2 2 + 32 h = ± √ −4 + 9 = ± √ 5 = ±2.236units The angle (direction) is computed as tanθ = 3 −2 θ = tan−1 ( 3 −2 ) = −56.31◦ = 56.31◦ The vector 3 −2 can therefore be written in magnitude-direction form as (± √ 5, 56.31◦ )
Addition Rule of Probability
It simply refers to the likelihood of an event taking place whenever the occurrence of an event is uncertain. The probability of a single event can be calculated by dividing the number of successful trials of that event by the total number of trials.
Expected Value
When a large number of trials are performed for any random variable ‘X’, the predicted result is most likely the mean of all the outcomes for the random variable and it is known as expected value also known as expectation. The expected value, also known as the expectation, is denoted by: E(X).
Probability Distributions
Understanding probability is necessary to know the probability distributions. In statistics, probability is how the uncertainty of an event is measured. This event can be anything. The most common examples include tossing a coin, rolling a die, or choosing a card. Each of these events has multiple possibilities. Every such possibility is measured with the help of probability. To be more precise, the probability is used for calculating the occurrence of events that may or may not happen. Probability does not give sure results. Unless the probability of any event is 1, the different outcomes may or may not happen in real life, regardless of how less or how more their probability is.
Basic Probability
The simple definition of probability it is a chance of the occurrence of an event. It is defined in numerical form and the probability value is between 0 to 1. The probability value 0 indicates that there is no chance of that event occurring and the probability value 1 indicates that the event will occur. Sum of the probability value must be 1. The probability value is never a negative number. If it happens, then recheck the calculation.
A. Identify the misconceptions, or problems in the provided solutions
if any and
B. Provide at two(2) remedies to the problems or/and misconceptions identified.
b. Suppose we choose an integer from the set of the first ten 10
positive integers. The
S = 1, 2, 3, ..., 10
Each of the sets
A = {2, 4, 6, 8, 10},
B = {1, 3, 5, 7, 9},
C = {3, 6, 9},
is an event. A is the event of choosing an even integer, B is the
event of choosing an odd integer while C is the event of choosing
an integer divisible by 3. The complement of C is is given by
C
0 = {1, 2, 4, 5, 7, 8, 10}.
The union of B and C is
B ∪ C = {1, 3, 3, 5, 6, 7, 9, 9}
and the number of elements in set B ∪ C is n(B ∪ C) = 8.
The intersection of B and C is B ∩ C = {3, 3, 9, 9}
and the number of elements in set B ∩ C is n(B ∪ C) = 4.
Notice that
(i) C ∪ C
0 = S
(ii) C ∩ C
0 = A ∪ B
The events B and C are mutually exclusive.
b. We want to draw a diagram to show the
write it in magnitude-direction form.
3
−2
The magnitude of the vector is
(hypotenuse)2 = (opposite)2 + (adjacent)2
h
2 = −2
2 + 32
h = ±
√
−4 + 9
= ±
√
5 = ±2.236units
The angle (direction) is computed as
tanθ =
3
−2
θ = tan−1
(
3
−2
)
= −56.31◦ = 56.31◦
The vector
3
−2
can therefore be written in magnitude-direction
form as
(±
√
5, 56.31◦
)
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