Respuesta :
Because the sectors are of equal size, each sector has the same probability of happening.
So for every number, you have a probability of 0.2 of it appearing.
Because you have three odd numbers, the probability to land on an odd number on each spin is 0.6. You can also say that the probability to not have an odd number is 0.4.
So what is the probability to have 0 odd number in two spins?
[tex]0.4 \times 0.4 = 0.16[/tex]
What is the probability to have 1 odd number in two spins? You can have one odd number on the first or one odd on the second spin, that is why we add the probabilities.
[tex]0.6 \times 0.4 + 0.4 \times 0.6 = 0.24 + 0.24 = 0.48[/tex]
What is the probability to have 2 odd numbers in two spins?
[tex]0.6 \times 0.6 = 0.36[/tex]
To verify if we did a good job, we add all the probabilities. We should get 1.
[tex]0.16+0.48+0.36 = 1[/tex].
So to slide your bars, you slide them up to the number I gave you for each case.
So for every number, you have a probability of 0.2 of it appearing.
Because you have three odd numbers, the probability to land on an odd number on each spin is 0.6. You can also say that the probability to not have an odd number is 0.4.
So what is the probability to have 0 odd number in two spins?
[tex]0.4 \times 0.4 = 0.16[/tex]
What is the probability to have 1 odd number in two spins? You can have one odd number on the first or one odd on the second spin, that is why we add the probabilities.
[tex]0.6 \times 0.4 + 0.4 \times 0.6 = 0.24 + 0.24 = 0.48[/tex]
What is the probability to have 2 odd numbers in two spins?
[tex]0.6 \times 0.6 = 0.36[/tex]
To verify if we did a good job, we add all the probabilities. We should get 1.
[tex]0.16+0.48+0.36 = 1[/tex].
So to slide your bars, you slide them up to the number I gave you for each case.
The probability distribution for the random variable X is evaluated to be:
- X= 0, P(X = 0) = 0.16
- X = 1, P(X = 1) = 0.48
- X = 2, P(X = 2) = 0.36
What is probability distribution?
Probability distribution of a random variable is the collection of value and its probability pair for values of that considered random variable.
It might be a function, a table etc.
How to calculate the probability of an event?
Suppose that there are finite elementary events in the sample space of the considered experiment, and all are equally likely.
Then, suppose we want to find the probability of an event E.
Then, its probability is given as
[tex]P(E) = \dfrac{\text{Number of favorable cases}}{\text{Number of total cases}} = \dfrac{n(E)}{n(S)}[/tex]
where favorable cases are those elementary events who belong to E, and total cases are the size of the sample space.
For this case, we have:
Spinner has values 1, 2, 3, 4, and 5
Spinner spun twice
- X = number of times an odd number comes.
- A = an odd number comes in first trial
- B = an odd number comes in second trial
A and B are independent events, so their simultaneous occurrence' probability is: [tex]P(A \cap B) = P(A)P(B)[/tex] (from definition independent events and chain rule).
Total odds' counts = 3 (they are 1, 3 and 5) = n(A) = n(B) (number of favorable events)
Total labels on spinner = 5 = n(S)
Thus, P(A) = n(A)/n(S) = P(B) = 3/5
X can be either 0 (no odd number in either of the spins), 1 (single times odd number as outcome), or 2 (both times odd comes).
Calculating the probabilities:
- Case 1: X = 0
Complement of A and complement of B needs to occur simultaneously.
[tex]P(X = 0) = P(A' \cap B') = P(\overline{A \cup B}) = 1 - P(A \cup B)[/tex]
By addition rule of probability, we get:
[tex]P(A \cup B) = P(A) + P(B) - P(A \cap B)\\\\P(A \cup B) = \dfrac{3}{5} + \dfrac{3}{5} - (\dfrac{3}{5})^2 = \dfrac{21}{25}[/tex]
Thus, [tex]P(X = 0) = 1 - P(A \cup B) = 1 - \dfrac{21}{25} = \dfrac{4}{25} = 0.16[/tex]
- Case 2: X = 1
Either A or B have to occur. That means it is sum of probability of (A and B' (complement of B))'s occurrence, and probability of (A' and B)'s occurrence
Thus, we get:
[tex]P(X = 1) = P(A \cap B') = P(A)P(B') +P(A' \cap B)\\P(X = 1)= P(A)(1-P(B)) + (1-P(A))P(B) \\P(X = 1)= \dfrac{3}{5}(1-\dfrac{3}{5}) + (1-\dfrac{3}{5})\dfrac{3}{5}\\\\P(X =1) = \dfrac{12}{25}[/tex]
- Case 3: X = 2
Both A and B has to occur simultaneously in this case. Thus,
[tex]P(X = 2) = P(A \cap B) = P(A)P(B) = (\dfrac{3}{5})^2 = \dfrac{9}{25} = 0.36[/tex]
Thus, the probability distribution for the random variable X is evaluated to be:
- X= 0, P(X = 0) = 0.16
- X = 1, P(X = 1) = 0.48
- X = 2, P(X = 2) = 0.36
Learn more about probability here:
brainly.com/question/1210781
