To determine the experimental probability that at least 2 out of 8 campers are from out of state, we need to analyze each number string and count how many 0's and 1's are in it. If there are 2 or more, we count it as a success. If there are fewer than 2, we count it as a failure.
Out of the 20 number strings given, the following have at least 2 0's or 1's:
- 69531217
- 89542756
- 89001254
- 01346895
- 12468503
- 20312346
- 79564328
- 59868542
- 75891003
- 56103249
- 58630126
- 57498510
- 76134860
- 52974168
- 03164985
- 03289642
- 94210145
- 70215789
That's a total of 18 successful number strings out of 20. Therefore, the experimental probability that in a cabin of 8 campers, at least 2 are from out of state is:
18/20 = 9/10 = Start Fraction 9 over 10 End Fraction
So the answer is (C) start fraction 7 over 10 end fraction.
At a summer camp, 20% of the students come from out of state. Megan wants to simulate cabins of 8 campers. She generates random digits from 0 to 9 and lets the digits 0 and 1 represent a camper from out of state. The table below shows her results.
69531217
89542756
89001254
01346895
12468503
20312346
79564328
59868542
75891003
56103249
58630126
57498510
76134860
52974168
03164985
03289642
68533236
94210145
70215789
32605254
What is the experimental probability that in a cabin of 8 campers, at least 2 are from out of state?
A. start fraction 7 over 20 end fraction
B. Start Fraction 13 over 20 End Fraction
C. start fraction 7 over 10 end fraction
D. four-fifths..
1 answer
1 answer