Food & Drink1 min ago
So My Math Practice Changed And I Actually Need Help With These Instead!
19 Answers
1.A number cube is rolled 120 times. The number 4 comes up 47 times. What is the experimental probability of rolling a 4? What is the theoretical probability of rolling a 4?
A)47/120;1/30
B)47/120;1/6
C)4/47;1/6
D)1/6;47/120
2.Food Express is running a special promotion in which customers can win a free gallon of milk with their food
purchase if there is a star on their receipt. So far, 147 of the first 156 customers have not received a star on their receipts. What is the experimental probability of winning a free gallon of milk?
A)11/156
B)49/52
C)2/39
D)3/52
3.A bag containing 5 green marbles, 8 red marbles, 11 orange marbles, 7 brown marbles, and 12 blue marbles. You choose a marble, replace it, and choose again. What is P(red, then blue)?
A)20/43
B)40/43
C)20/1849
D)96/1849
4.You and 5 friends go to a concert. In how many different ways can you sit in the assigned seats?
A)720
B)120
C)30
D)15
A)47/120;1/30
B)47/120;1/6
C)4/47;1/6
D)1/6;47/120
2.Food Express is running a special promotion in which customers can win a free gallon of milk with their food
purchase if there is a star on their receipt. So far, 147 of the first 156 customers have not received a star on their receipts. What is the experimental probability of winning a free gallon of milk?
A)11/156
B)49/52
C)2/39
D)3/52
3.A bag containing 5 green marbles, 8 red marbles, 11 orange marbles, 7 brown marbles, and 12 blue marbles. You choose a marble, replace it, and choose again. What is P(red, then blue)?
A)20/43
B)40/43
C)20/1849
D)96/1849
4.You and 5 friends go to a concert. In how many different ways can you sit in the assigned seats?
A)720
B)120
C)30
D)15
Answers
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The experimental probability is based on the results of the experiment ONLY and you ignore everything else.
The theoretical probbility is what should have happened and you ignore the results of the experiment.
Therefore
1. In the experiment as the number 4 was thrown in 47 out of 120 times, the experimental probability = 47/120
Theoretical probability now. 4 is just one number on the die and there are 6 numbers (1,2,3,4,5,6). Therefore the probability that a 4 occurs = 1/6
Answer = B
2. As there are 156 people surveyed and 147 have not won, it means that 156 - 147 = 9 HAVE won.
The experimental probability of winning is therefore 9/156 which then simplifies by dividing top and bottom by 3 to 3/52
Answer = D
Now I shall leave you to try the last 2. Answer this question yourself with your answers and I shall check them and correct them if needed!
The theoretical probbility is what should have happened and you ignore the results of the experiment.
Therefore
1. In the experiment as the number 4 was thrown in 47 out of 120 times, the experimental probability = 47/120
Theoretical probability now. 4 is just one number on the die and there are 6 numbers (1,2,3,4,5,6). Therefore the probability that a 4 occurs = 1/6
Answer = B
2. As there are 156 people surveyed and 147 have not won, it means that 156 - 147 = 9 HAVE won.
The experimental probability of winning is therefore 9/156 which then simplifies by dividing top and bottom by 3 to 3/52
Answer = D
Now I shall leave you to try the last 2. Answer this question yourself with your answers and I shall check them and correct them if needed!
Actually - as they are totally different, I shall answer them anyway, it is just that I like to see somebody try first. I would rather correct an incorrect answer than do the work myself.
3. There are 43 marbles in the bag (5+8+11+7+12) and of these 8 are red. Therefore the probability of getting a red first = 8/43
Then as you replace the red, there remain 43 marbles in the bag and of these 12 are blue. Therefore the probability of getting a blue second = 12/43
Whenever you have MULTIPLE events that all MUST occur you MULTIPLY the individual probabilities.
Therefore P(red then blue) = 8/43 * 12/43 = 96/1849
= Answer D
4. Straight forward permutations = 5! = 5 * 4 * 3 * 2 * 1 = 120
Answer B
The first seat can be occupied by any of 5
The second seat can be occupied by any of 4 (one already in the first seat)
The third seat can be occupied by any of 3. (two lredy in sets 1&2)
the 4th seat by either of 2
The last seat by whoever is left over = 1
5x4x3x2x1 = 120
3. There are 43 marbles in the bag (5+8+11+7+12) and of these 8 are red. Therefore the probability of getting a red first = 8/43
Then as you replace the red, there remain 43 marbles in the bag and of these 12 are blue. Therefore the probability of getting a blue second = 12/43
Whenever you have MULTIPLE events that all MUST occur you MULTIPLY the individual probabilities.
Therefore P(red then blue) = 8/43 * 12/43 = 96/1849
= Answer D
4. Straight forward permutations = 5! = 5 * 4 * 3 * 2 * 1 = 120
Answer B
The first seat can be occupied by any of 5
The second seat can be occupied by any of 4 (one already in the first seat)
The third seat can be occupied by any of 3. (two lredy in sets 1&2)
the 4th seat by either of 2
The last seat by whoever is left over = 1
5x4x3x2x1 = 120
The first one doesn't need an image. Here are the questions:19. A box contains 95 pink rubber bands and 90 brown rubber bands. You select a rubber band at random from the box. Find each probability. Write the probability as a fraction in simplest form.
a. Find the theoretical probability of selecting a pink rubber band.
b. Find the theoretical probability of selecting a brown rubber band.
c. You repeatedly choose a rubber band from the box, record the color, and put the rubber band back in the box. The results are shown in the table below. Find the experimental probability of each color based on the table:
Outcome Occurrences
Pink 36
Brown 33
20. The diagram below shows the contents of a jar from which you select marbles at random.
a. What is the probability of selecting a red marble, replacing it, and then selecting a blue marble?
b. What is the probability of selecting a red marble, setting it aside, and then selecting a blue marble
c. Are the answers to part a and part b the same? Why or why not?
a. Find the theoretical probability of selecting a pink rubber band.
b. Find the theoretical probability of selecting a brown rubber band.
c. You repeatedly choose a rubber band from the box, record the color, and put the rubber band back in the box. The results are shown in the table below. Find the experimental probability of each color based on the table:
Outcome Occurrences
Pink 36
Brown 33
20. The diagram below shows the contents of a jar from which you select marbles at random.
a. What is the probability of selecting a red marble, replacing it, and then selecting a blue marble?
b. What is the probability of selecting a red marble, setting it aside, and then selecting a blue marble
c. Are the answers to part a and part b the same? Why or why not?
Here is the link to the second one, let me know if you can see it or not: https:/
19 a = the number of pink bands / (the number of pink bands + the number of brown bands)
b = the number of brown / (the number of pink + the number of brown)
c = You have done an experiment 36 + 33 = 69 times
In these experiments you obtained pink 36 times
Experimental probability pink = 36 /69 = 12/23
Similarly experimental probability brown = 33/69 = 11/23
b = the number of brown / (the number of pink + the number of brown)
c = You have done an experiment 36 + 33 = 69 times
In these experiments you obtained pink 36 times
Experimental probability pink = 36 /69 = 12/23
Similarly experimental probability brown = 33/69 = 11/23
20 - access denied but no problem because I shall explain what you do
First draw P(red) = number of red marbles / total number of marbles in the bag.
If replaced then the number of marbles in the bag remains the same
P(second = blue) = the number of blue marbles in the bag / total marbles in the bag
Now multiply the individul probabilities together
P(red then blue) = P(first Red) x P(second blue) =
b) P(first red) = exactly the same as above
Now this is where it changes
Say there were originally 40 marbles in the bag, after one has been removed there will now only be 39 left to draw the second blue from.
Therefore rather than number of blue /40, it will be number of blue divided by 39
Again multiply the individual probabilities together = R/40 * B/39
c) They are different and I hope you can now explain why!
First draw P(red) = number of red marbles / total number of marbles in the bag.
If replaced then the number of marbles in the bag remains the same
P(second = blue) = the number of blue marbles in the bag / total marbles in the bag
Now multiply the individul probabilities together
P(red then blue) = P(first Red) x P(second blue) =
b) P(first red) = exactly the same as above
Now this is where it changes
Say there were originally 40 marbles in the bag, after one has been removed there will now only be 39 left to draw the second blue from.
Therefore rather than number of blue /40, it will be number of blue divided by 39
Again multiply the individual probabilities together = R/40 * B/39
c) They are different and I hope you can now explain why!
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