At 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:
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. 7/20
B. 13/20
C. 7/10
D. 4/5
Is it B?
What is the experimental probability that in a cabin of 8 campers, none is from out of state?
A. 1/20
B. 1/10
C. 1/5 **?
D. 1/4
1. C
2. D
3. B
4. C
C
D
B
C
For question one, you need to look at each number (of eight digits) which represents 8 campers. Any digit 0 or 1 represents out-of-state campers.
Count how many of the 20 numbers contain at least two of 0 or 1.
For example, the first number is 69531217
which means that there are two (satisfies "at least two") out-of-state campers. So it counts as one success out of the 20 "experiments". Repeat for the other 19 numbers and give the count, divided by 20 experiments as the count.
(2) For this, you have to count (as success) the number of numbers with no 0 and no 1. (example: 89542756).
Do the same for all 20 numbers and count the number of successes.
Ms. Sue? Reed? Can someone help me? Thank you!
So... are those answers I got right?
Yes, they are correct!
I didn't notice you gave answers, sorry.
Thank You, Your Mom :) <3
1. C
2. D
3. B
4. C
100%
Thanks guys!
To solve this problem, we need to count the number of cabins where at least 2 campers are from out of state, and the number of cabins where none of the campers are from out of state. Then, we can calculate the experimental probability by dividing the number of successful outcomes by the total number of outcomes.
Let's start by counting the number of cabins where at least 2 campers are from out of state. Looking at the given table:
- In the first cabin (69531217), there are 2 campers from out of state (0 and 1).
- In the second cabin (89542756), there are 3 campers from out of state (1, 0, and 1).
- In the third cabin (89001254), there are 2 campers from out of state (0 and 0).
- In the fourth cabin (01346895), there are 4 campers from out of state (0, 1, 0, and 1).
- In the fifth cabin (12468503), there are 2 campers from out of state (1 and 0).
- ... (continue counting for all cabins)
By counting, you will find that there are 12 cabins where at least 2 campers are from out of state.
Next, let's count the number of cabins where none of the campers are from out of state:
- In the first cabin (69531217), there is at least 1 camper from out of state (0 or 1).
- In the second cabin (89542756), there is at least 1 camper from out of state (1, 0, or 1).
- In the third cabin (89001254), there is at least 1 camper from out of state (0 or 0).
- In the fourth cabin (01346895), there is at least 1 camper from out of state (0, 1, 0, or 1).
- In the fifth cabin (12468503), there is at least 1 camper from out of state (1 or 0).
- ... (continue counting for all cabins)
By counting, you will find that there are 0 cabins where none of the campers are from out of state.
Now, let's calculate the experimental probability for each case:
- The experimental probability that at least 2 campers are from out of state is 12 (number of successful outcomes) divided by 20 (total number of outcomes), which simplifies to 3/5.
- The experimental probability that none of the campers are from out of state is 0 (number of successful outcomes) divided by 20 (total number of outcomes), which simplifies to 0.
Therefore, the correct answer is:
- For the experimental probability that at least 2 campers are from out of state: D. 4/5
- For the experimental probability that none of the campers are from out of state: A. 1/20