To find the value of x, we know that the sum of all probabilities must equal 1 in a probability model.
0.25 + 0.18 + 0.30 + 0.17 + x = 1
Solving for x:
0.90 + x = 1
x = 1 - 0.90
x = 0.10
Therefore, the value of x is 0.10.
If the table below represents 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 (1 point)
Responses
1
1
0.10
0.10
0.20
0.20
0.01
9 answers
Use the table to answer the question.
Number of White Balls Selected Number of Red Balls Selected
Bowl A 45 55
Bowl B 60 40
Bowls A and B contain a number of white and red balls. Clark repeatedly 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 difference in the expected number of white and red balls in Bowl B?
(1 point)
Responses
50
50
200
200
100
100
300
Number of White Balls Selected Number of Red Balls Selected
Bowl A 45 55
Bowl B 60 40
Bowls A and B contain a number of white and red balls. Clark repeatedly 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 difference in the expected number of white and red balls in Bowl B?
(1 point)
Responses
50
50
200
200
100
100
300
To find the estimated difference in the expected number of white and red balls in Bowl B, we can calculate the expected number of white balls and red balls selected from Bowl B based on the given proportions.
In Bowl B, the proportion of white balls is 60/100 = 0.6
The expected number of white balls selected from Bowl B = 0.6 * 500 = 300
Similarly, the proportion of red balls in Bowl B is 40/100 = 0.4
The expected number of red balls selected from Bowl B = 0.4 * 500 = 200
The estimated difference in the expected number of white and red balls in Bowl B:
300 (expected white balls) - 200 (expected red balls) = 100
Therefore, the estimated difference in the expected number of white and red balls in Bowl B is 100.
In Bowl B, the proportion of white balls is 60/100 = 0.6
The expected number of white balls selected from Bowl B = 0.6 * 500 = 300
Similarly, the proportion of red balls in Bowl B is 40/100 = 0.4
The expected number of red balls selected from Bowl B = 0.4 * 500 = 200
The estimated difference in the expected number of white and red balls in Bowl B:
300 (expected white balls) - 200 (expected red balls) = 100
Therefore, the estimated difference in the expected number of white and red balls in Bowl B is 100.
What is the probability of rolling an odd number on the first roll of a six-sided cube and rolling an even number on the second roll?(1 point)
Responses
2
2
1
1
12
Start Fraction 1 over 2 End Fraction
14
Responses
2
2
1
1
12
Start Fraction 1 over 2 End Fraction
14
To find the probability of rolling an odd number on the first roll of a six-sided cube and rolling an even number on the second roll, we need to consider the total possible outcomes of each roll.
1. Probability of rolling an odd number on the first roll:
There are 3 odd numbers on a six-sided cube (1, 3, 5), out of a total of 6 numbers.
Therefore, the probability of rolling an odd number on the first roll is 3/6 = 1/2.
2. Probability of rolling an even number on the second roll:
There are 3 even numbers on a six-sided cube (2, 4, 6), out of a total of 6 numbers.
Therefore, the probability of rolling an even number on the second roll is 3/6 = 1/2.
To find the combined probability of both events occurring (rolling an odd number followed by an even number), we multiply the individual probabilities:
P(odd on first roll) * P(even on second roll) = (1/2) * (1/2) = 1/4
Therefore, the probability of rolling an odd number on the first roll of a six-sided cube and rolling an even number on the second roll is 1/4.
1. Probability of rolling an odd number on the first roll:
There are 3 odd numbers on a six-sided cube (1, 3, 5), out of a total of 6 numbers.
Therefore, the probability of rolling an odd number on the first roll is 3/6 = 1/2.
2. Probability of rolling an even number on the second roll:
There are 3 even numbers on a six-sided cube (2, 4, 6), out of a total of 6 numbers.
Therefore, the probability of rolling an even number on the second roll is 3/6 = 1/2.
To find the combined probability of both events occurring (rolling an odd number followed by an even number), we multiply the individual probabilities:
P(odd on first roll) * P(even on second roll) = (1/2) * (1/2) = 1/4
Therefore, the probability of rolling an odd number on the first roll of a six-sided cube and rolling 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 randomly selected without replacement. What is the probability of selecting two carnations? Express the answer in the simplest form of fraction.(1 point)
To calculate the probability of selecting two carnations from the vase, we first need to find the total number of ways to choose 2 flowers from the total number of flowers in the vase, and then determine the number of ways to choose 2 carnations from the total number of carnations in the vase.
Total number of ways to choose 2 flowers from 5 begonias, 6 carnations, and 7 tulips:
Total number of flowers = 5 (begonias) + 6 (carnations) + 7 (tulips) = 18 flowers
Total number of ways to choose 2 flowers out of 18 = 18C2 = (18 * 17) / (2 * 1) = 153 ways
Number of ways to choose 2 carnations out of 6 carnations:
Number of ways to choose 2 carnations = 6C2 = (6 * 5) / (2 * 1) = 15 ways
Probability of selecting two carnations = Number of ways to choose 2 carnations / Total number of ways to choose 2 flowers
Probability = 15 / 153
Probability = 5 / 51
Therefore, the probability of selecting two carnations is 5/51.
Total number of ways to choose 2 flowers from 5 begonias, 6 carnations, and 7 tulips:
Total number of flowers = 5 (begonias) + 6 (carnations) + 7 (tulips) = 18 flowers
Total number of ways to choose 2 flowers out of 18 = 18C2 = (18 * 17) / (2 * 1) = 153 ways
Number of ways to choose 2 carnations out of 6 carnations:
Number of ways to choose 2 carnations = 6C2 = (6 * 5) / (2 * 1) = 15 ways
Probability of selecting two carnations = Number of ways to choose 2 carnations / Total number of ways to choose 2 flowers
Probability = 15 / 153
Probability = 5 / 51
Therefore, the probability of selecting two carnations is 5/51.
A standard deck of 52 cards contains four suits: hearts, diamonds, clubs, and spades. Each suit has 13 cards: ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, jack, queen, and king. Two cards are randomly drawn without replacement. Calculate the probability of drawing two diamond cards. Express your answer in percent form rounding to the nearest hundredth.(1 point)
%
%
To find the probability of drawing two diamond cards from a standard deck of 52 cards, we first need to determine the total number of ways to draw 2 cards from the deck and the number of ways to draw 2 diamond cards.
1. Total number of ways to draw 2 cards from 52:
The total number of ways to draw 2 cards from 52 cards is given by the combination formula: 52C2 = (52 * 51) / (2 * 1) = 1326.
2. Number of ways to draw 2 diamond cards from 13 diamonds:
The number of ways to draw 2 diamond cards from 13 diamonds is given by the combination formula: 13C2 = (13 * 12) / (2 * 1) = 78.
3. Probability of drawing two diamond cards:
Probability = Number of ways to draw 2 diamond cards / Total number of ways to draw 2 cards
Probability = 78 / 1326
Probability ≈ 0.0588
Converting the probability to a percentage and rounding to the nearest hundredth:
Probability ≈ 0.0588 * 100 ≈ 5.88%
Therefore, the probability of drawing two diamond cards from a standard deck is approximately 5.88%.
1. Total number of ways to draw 2 cards from 52:
The total number of ways to draw 2 cards from 52 cards is given by the combination formula: 52C2 = (52 * 51) / (2 * 1) = 1326.
2. Number of ways to draw 2 diamond cards from 13 diamonds:
The number of ways to draw 2 diamond cards from 13 diamonds is given by the combination formula: 13C2 = (13 * 12) / (2 * 1) = 78.
3. Probability of drawing two diamond cards:
Probability = Number of ways to draw 2 diamond cards / Total number of ways to draw 2 cards
Probability = 78 / 1326
Probability ≈ 0.0588
Converting the probability to a percentage and rounding to the nearest hundredth:
Probability ≈ 0.0588 * 100 ≈ 5.88%
Therefore, the probability of drawing two diamond cards from a standard deck is approximately 5.88%.
3 answers