An experiment involves picking a card from a selecting of cars numbered 5,10,15. In equation form, what is the probability model for this experiment. Lost the value of x in ascending order
F(x) = ( ) where x = ( ) ( ) ( )
F(x) = (1/3) where x = 5, 10, 15
If the table below represent a probability model based on observed frequencies what is the value of x
Y. C. A. R. E. S
P (Y) 0.25. 0.18. 0.30. 0.17. X
A. 0.01
B. 1
C. 0.10
D. 0.20
To find the value of X, we know that the sum of all probabilities must equal 1.
Given:
P(C) = 0.18
P(A) = 0.30
P(R) = 0.17
P(E) = X
Sum of probabilities = P(Y) + P(C) + P(A) + P(R) + P(E) = 0.25 + 0.18 + 0.30 + 0.17 + X
0.25 + 0.18 + 0.30 + 0.17 + X = 1
0.90 + X = 1
X = 1 - 0.90
X = 0.10
Therefore, the value of X is 0.10, so the answer is C. 0.10.
Use the table to answer the questions
Number of white balls selected
45
60
—————
Number of red balls selected
55
40
Bowls A and B contain a number of white and red balls. Class repeated selected a ball from both bowls and recorded the results in a table. If there are 500 balls in bowl B, what is the estimated sufferance in the expected number is white abcs red balls in bowl B
To find the estimated distribution of white and red balls in Bowl B, we will use the data provided in the table.
From the table:
Number of white balls selected from Bowl B = 60
Number of red balls selected from Bowl B = 40
Total number of balls selected from Bowl B = 60 (white) + 40 (red) = 100
Given that there are 500 balls in Bowl B, we can set up a proportion to estimate the distribution of white and red balls in Bowl B:
White balls in Bowl B / Total balls in Bowl B = Number of white balls selected / Total number of balls selected from Bowl B
Let x be the estimated number of white balls in Bowl B:
x/500 = 60/100
x = 500 * (60/100)
x = 300
Therefore, the estimated number of white balls in Bowl B is 300.
Similarly, to find the estimated number of red balls in Bowl B:
Total red balls in Bowl B = 500 - 300 (white balls) = 200
Therefore, the estimated number of red balls in Bowl B is 200.
So, the estimated distribution of white and red balls in Bowl B is 300 white balls and 200 red balls.
A. 100
B 300
C. 50
D. 200
To estimate the distribution of white and red balls in Bowl B based on the data provided in the table:
Number of white balls selected from Bowl B = 60
Number of red balls selected from Bowl B = 40
Total number of balls selected from Bowl B = 60 (white) + 40 (red) = 100
Given that there are 500 balls in Bowl B, we can estimate the number of white balls in proportion to the total number of balls:
Estimated number of white balls in Bowl B = (60/100) * 500
Estimated number of white balls in Bowl B = 300
Therefore, the estimated number of white balls in Bowl B is 300.
Since the total number of balls is 500, we can find the estimated number of red balls in Bowl B by subtracting the estimated number of white balls from the total:
Estimated number of red balls in Bowl B = 500 - 300
Estimated number of red balls in Bowl B = 200
Therefore, the estimated distribution of white and red balls in Bowl B is 300 white balls and 200 red balls.
The correct option is:
B. 300
What is the probability of rolling an odd number in the first will of a six sided cube and rolling an even number in the 2nd doll
To find the probability of rolling an odd number on the first roll of a six-sided die and an even number on the second roll, we need to calculate the individual probabilities and then multiply them together.
1. Probability of rolling an odd number on the first roll:
In a standard six-sided die, there are 3 odd numbers (1, 3, 5) out of the total 6 numbers.
Probability of rolling an odd number on the first roll = Number of favorable outcomes / Total number of outcomes
Probability of rolling an odd number on the first roll = 3/6 = 1/2
2. Probability of rolling an even number on the second roll:
There are 3 even numbers (2, 4, 6) out of the total 6 numbers.
Probability of rolling an even number on the second roll = Number of favorable outcomes / Total number of outcomes
Probability of rolling an even number on the second roll = 3/6 = 1/2
Now, to find the probability of both events occurring, we multiply the individual probabilities together:
Probability of rolling an odd number on the first roll and an even number on the second roll = Probability of first event * Probability of second event
Probability of both events = (1/2) * (1/2) = 1/4
Therefore, the probability of rolling an odd number on the first roll and an even number on the second roll is 1/4.
A big flower vase has 5 begonias, 6 carnations, and 7 tulips. Two flowers are easily selected without replacement. What is the probability of selecting two carnations. Express the answer in the simplest form of fractions
To find the probability of selecting two carnations from the big flower vase containing 5 begonias, 6 carnations, and 7 tulips, we first need to calculate the total number of ways to select any two flowers from the vase.
Total number of flowers in the vase = 5 begonias + 6 carnations + 7 tulips = 18 flowers
The total number of ways to select 2 flowers from 18 flowers is given by the combination formula:
Total number of ways to select 2 flowers out of 18 = C(18, 2) = 18! / [(18-2)! * 2!] = 153 ways
Now, to find the number of ways to select 2 carnations out of the 6 available:
Number of ways to select 2 carnations = C(6, 2) = 6! / [(6-2)! * 2!] = 15 ways
Therefore, the probability of selecting two carnations from the vase is given by the number of ways to select 2 carnations divided by the total number of ways to select any 2 flowers:
Probability = Number of ways to select 2 carnations / Total number of ways to select 2 flowers
Probability = 15 / 153 = 5 / 51
So, the probability of selecting two carnations is 5/51.