How does the experimental probability of rolling a 3 compare with the theoretical probability of rolling a 3
2 answers:
You might be interested in
Answer:
81 * 4 = 324.
Step-by-step explanation:
There are 81 * 4 = 324 such integers.
There are 9000 four-digit integers from 1000 to 9999. Let’s refer to the digit that occurs three times as the triplet and the digit that occurs once as the singleton.
In the first (most significant) position there are 9 possible values for the singleton, 1 through 9. In each case there are also 9 possible values for the triplet. For example if the singleton was a 1, the triplets could be anything but a 1: 1000, 1222, 1333, …, 1999. So there are 9 * 9 = 81 cases with the singleton in the first position.
A singleton in the second position can have 10 possible values, 0 through 9. If it’s a 0 there are 9 possible values for the triplet: 1011, 2022, 3033, …, 9099. If it’s anything else there are only 8 possible values for the triplet since the triplet cannot be 0: 2122, 3133, 4144, …, 9199. So there are 9 + (8 * 9) = 81 cases with the singleton in the second position.
The third and fourth positions work just like the second. So the grand total is 81 * 4 = 324.
If you plot the 9000 four-digit integers as a 100 x 90 grid of squares and color the integers with three same digits in red, an interesting pattern occurs:
There are 81 * 4 = 324 such integers.
There are 9000 four-digit integers from 1000 to 9999. Let’s refer to the digit that occurs three times as the triplet and the digit that occurs once as the singleton.
In the first (most significant) position there are 9 possible values for the singleton, 1 through 9. In each case there are also 9 possible values for the triplet. For example if the singleton was a 1, the triplets could be anything but a 1: 1000, 1222, 1333, …, 1999. So there are 9 * 9 = 81 cases with the singleton in the first position.
A singleton in the second position can have 10 possible values, 0 through 9. If it’s a 0 there are 9 possible values for the triplet: 1011, 2022, 3033, …, 9099. If it’s anything else there are only 8 possible values for the triplet since the triplet cannot be 0: 2122, 3133, 4144, …, 9199. So there are 9 + (8 * 9) = 81 cases with the singleton in the second position.
The third and fourth positions work just like the second. So the grand total is 81 * 4 = 324.
If you plot the 9000 four-digit integers as a 100 x 90 grid of squares and color the integers with three same digits in red, an interesting pattern occurs:
(If you zoom in, the four-digit number is printed inside each red square - not sure whether or not that will come through after Quora processes the image.)
Hope this helps!
Brain-List?
Answer:
41.92%
Step-by-step explanation:
In table attached standard normal distribution curve is shown.
Data
mean, µ = 4.3 minutes
standard deviation, σ = 0.5 minutes
desired point, x = 5 minutes
x-axis, Z = (x - µ)/σ
The probability that calls last between 4.3 and 5.0 minutes is equivalent to find the area below the curve between Z = 0 and Z = (5 - 4.3)/0.5 = 1.4, which is 0.4192, or 41.92%
