Anthony observes that when people are leaving class, about 50% turn left and about 50% turn right. Anthony wants to estimate the probability that fewer than 3 of the first 4 students will turn left when leaving class tomorrow.
To do this, he lets H represent turning left when leaving class and T represent turning right. He then flips a coin 4 times, recording the results. He repeats this process for a total of 20 trials.
HHTT, THHT, HTTT, HTHH,HHTT, HTHT, THTH, TTHH,HHHH, HTTH, HHHT, TTHT,THHH, THTH, HHTT, HTTT,HTTH, HHTT, HTHH, HHTH
Drag and drop the correct fraction to the box to estimate the probability, using this simulation.
Put responses in the correct input to answer the question. Select a response, navigate to the desired input and insert the response. Responses can be selected and inserted using the space bar, enter key, left mouse button or touchpad. Responses can also be moved by dragging with a mouse.
Estimated probability that fewer than 3 of the first 4 students leaving class tomorrow will turn left:
To estimate the probability that fewer than 3 of the first 4 students will turn left when leaving class tomorrow, we need to count the number of trials in which fewer than 3 heads (H) appear in the first 4 flips.
Out of the 20 trials, we can see that there are 11 trials where fewer than 3 heads appear:
HHTT, THHT, HTTT, HTHT, THTH, TTHH, HTTH, TTHT, THHH, THTH, and HHTT
So the estimated probability is 11/20.
the answer choices are 1/20 3/10 7/10 19/20
Apologies for the confusion. Upon reviewing the trials again, we can see that there are 11 trials where fewer than 3 heads appear. Therefore, the estimated probability that fewer than 3 of the first 4 students leaving class tomorrow will turn left is 11/20.
Casey finds that when asked for their email address, 15 of her customers will share their address with her.
To simulate the behavior of the next 4 customers, she writes the numbers 1, 2, 3, 4, and 5 on cards and mixes them up. She then picks 4 cards, records the results, and puts the cards back. She repeats this experiment for a total of 20 trials.
The results are shown in this list. Here, she let 1 represent someone sharing an email address and 2, 3, 4, and 5 represent someone not sharing an email address.
4121, 1223, 4233, 3344, 2434, 4241, 5415, 3421, 5214, 2234,4211, 1432, 5532, 4114, 4325, 4454, 4523, 4432, 3423, 1352
Using this simulation, what is the probability that, out of the next 4 customers, fewer than 2 will share their email address?
Enter your answer, as a fraction in simplified form, in the box.
To determine the probability that, out of the next 4 customers, fewer than 2 will share their email address, we need to count the number of trials in which fewer than 2 instances of "1" (indicating someone sharing their email address) appear in the 4 trials.
Out of the 20 trials listed, we can see that there are 12 trials where fewer than 2 instances of "1" appear:
1223, 3344, 2434, 4241, 3421, 2234, 4211, 1432, 5532, 4325, 4432, 3423
Therefore, the probability is 12/20, which simplifies to 3/5.
A company says that the probability of finding an all-star card in a game pack is 12.5%. A student wants to determine the probability that, if he buys 6 game packs, 2 or more will have an all-star card.
To do this, he uses a random number table to pick 6 numbers from 1 to 8. He lets 1 represent a pack with an all-star card and 2 through 8 represent a pack without an all-star card. He repeats this for a total of 15 trials. The results are shown in the table.
561673 366468 314255 361362 838443
126285 257231 768551 652585 468113
123711 322764 487626 783134 384664
Select from the drop-down menus to correctly complete each statement.
To estimate the probability that 2 or more out of 6 packs will have an all-star card, the student must count the number of trials with
Choose...
and divide by the number of trials.
This gives an estimated probability of about
Choose...
.
To estimate the probability that 2 or more out of 6 packs will have an all-star card, the student must count the number of trials with 2 or more instances of "1" (packs with an all-star card) and divide by the number of trials.
Upon examining the table, we can see that there are 6 trials with 2 or more instances of "1" :
561673, 361362, 126285, 257231, 652585, 322764
Therefore, the estimated probability is 6/15, which simplifies to 2/5.
in decimal?
To convert the estimated probability into decimal form, we divide the numerator (2) by the denominator (5):
2 / 5 = 0.4
Therefore, the estimated probability is approximately 0.4.