A. 75
Explanation: If the probability of selecting a shopper who prefers plastic bags is 50%, then out of 150 shoppers, we can expect that 50% of them, or 75 shoppers, will prefer plastic bags.
Probability Unit Test, Unit 8, Lesson 10, 7th grade
Suppose the probability of selecting a supermarket shopper who prefers plastic bags instead of paper bags is 50%. Out of 150 shoppers, how any can you expect will prefer plastic bags?
A. 75
B. 50
C. 100
D. 70
11 answers
A single coin is tossed 300 times. Heads were observed 180 times. What is the long-run relative frequency of tails? Express the answer in decimal form.
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To find the long-run relative frequency of tails, we first need to find the number of times tails were observed.
Total number of tosses = 300
Number of times heads were observed = 180
Number of times tails were observed = Total number of tosses - Number of times heads were observed
Number of times tails were observed = 300 - 180
Number of times tails were observed = 120
Now, we can find the relative frequency of tails by dividing the number of times tails were observed by the total number of tosses:
Relative frequency of tails = Number of times tails were observed / Total number of tosses
Relative frequency of tails = 120 / 300
Relative frequency of tails = 0.4
Therefore, the long-run relative frequency of tails is 0.4.
Total number of tosses = 300
Number of times heads were observed = 180
Number of times tails were observed = Total number of tosses - Number of times heads were observed
Number of times tails were observed = 300 - 180
Number of times tails were observed = 120
Now, we can find the relative frequency of tails by dividing the number of times tails were observed by the total number of tosses:
Relative frequency of tails = Number of times tails were observed / Total number of tosses
Relative frequency of tails = 120 / 300
Relative frequency of tails = 0.4
Therefore, the long-run relative frequency of tails is 0.4.
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
A. 0.10
B. 1
C. 0.01
D. 0.20
Y | C | A | R | E | S
P(Y)|0.25 | 0.18| 0.30 | 0.17| x
A. 0.10
B. 1
C. 0.01
D. 0.20
To find the value of x, we know that the sum of all probabilities in a probability model must equal 1.
Given the table:
0.25 + 0.18 + 0.30 + 0.17 + x = 1
Adding the given probabilities:
0.90 + x = 1
Subtracting 0.90 from both sides:
x = 1 - 0.90
x = 0.10
Therefore, the value of x is 0.10, which corresponds to option A.
Given the table:
0.25 + 0.18 + 0.30 + 0.17 + x = 1
Adding the given probabilities:
0.90 + x = 1
Subtracting 0.90 from both sides:
x = 1 - 0.90
x = 0.10
Therefore, the value of x is 0.10, which corresponds to option A.
White | Red
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?
A. 200
B. 100
C. 50
D. 300
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?
A. 200
B. 100
C. 50
D. 300
First, we need to calculate the expected number of white and red balls in Bowl B based on the probabilities given in the table.
In Bowl B, the probability of selecting a white ball is 60/100 = 0.6
The expected number of white balls in Bowl B = 0.6 * 500 (total balls in Bowl B) = 300
Similarly, the expected number of red balls in Bowl B = 0.4 * 500 = 200
The estimated difference in the expected number of white and red balls in Bowl B:
300 (Expected number of white balls) - 200 (Expected number of red balls) = 100
Therefore, the estimated difference in the expected number of white and red balls in Bowl B is 100, which corresponds to option B.
In Bowl B, the probability of selecting a white ball is 60/100 = 0.6
The expected number of white balls in Bowl B = 0.6 * 500 (total balls in Bowl B) = 300
Similarly, the expected number of red balls in Bowl B = 0.4 * 500 = 200
The estimated difference in the expected number of white and red balls in Bowl B:
300 (Expected number of white balls) - 200 (Expected number of red balls) = 100
Therefore, the estimated difference in the expected number of white and red balls in Bowl B is 100, which corresponds to option B.
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?
A. 1/4
B. 1
C. 2
D. 1/2
A. 1/4
B. 1
C. 2
D. 1/2
When rolling a fair six-sided die, there are a total of 3 odd numbers (1, 3, 5) and 3 even numbers (2, 4, 6).
The probability of rolling an odd number on the first roll is 3/6 or 1/2.
The probability of rolling an even number on the second roll is also 3/6 or 1/2.
To find the probability of both events happening, we multiply the probabilities:
P(odd & even) = P(odd) * P(even) = 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, which corresponds to option A.
The probability of rolling an odd number on the first roll is 3/6 or 1/2.
The probability of rolling an even number on the second roll is also 3/6 or 1/2.
To find the probability of both events happening, we multiply the probabilities:
P(odd & even) = P(odd) * P(even) = 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, which corresponds to option A.
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.
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First, we need to determine the total number of ways of choosing 2 cards from the 52 cards in a standard deck.
The total number of ways to choose the first card is 52 since there are 52 cards in the deck.
After drawing the first diamond card, there are 12 remaining diamond cards out of 51 cards remaining in the deck.
Therefore, the probability of drawing the second diamond card is 12/51.
The probability of drawing two diamond cards is the product of the probabilities of drawing the first and second diamond cards:
P(diamond, diamond) = (13/52) * (12/51) = (1/4) * (4/17) = 4/68 = 1/17
To express this as a percentage, we multiply by 100:
1/17 * 100 ≈ 5.88
Therefore, the probability of drawing two diamond cards is approximately 5.88%.
The total number of ways to choose the first card is 52 since there are 52 cards in the deck.
After drawing the first diamond card, there are 12 remaining diamond cards out of 51 cards remaining in the deck.
Therefore, the probability of drawing the second diamond card is 12/51.
The probability of drawing two diamond cards is the product of the probabilities of drawing the first and second diamond cards:
P(diamond, diamond) = (13/52) * (12/51) = (1/4) * (4/17) = 4/68 = 1/17
To express this as a percentage, we multiply by 100:
1/17 * 100 ≈ 5.88
Therefore, the probability of drawing two diamond cards is approximately 5.88%.
8 answers