Dawn and abed each roll a standard die obtaining a random number from 1 to 6. if the probability that dawns number is larger tha
1 answer:
Answer:
a+b = 12
Step-by-step explanation:
There are 36 possible outcomes for rolling two dice. Let A be the number rolled by Abed and D be the number rolled by Dawn. The sample space, in the format [A,D], of all of the possibilities in which Dawn's number is larger is:
[5,6]
[4,5] [4,6]
[3.4] [3,5] [3.6]
[2,3] [2.4] [2,5] [2.6]
[1,2] [1,3] [1.4] [1,5] [1.6]
There are 15 ways out of 36 possible ways of Dawn getting the higher number. In the simplest form:
You might be interested in
Answer:
a) 51.60% probability that in a given year there will be less than 21 earthquakes.
b) 49.35% probability that in a given year there will be between 18 and 23 earthquakes.
Step-by-step explanation:
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
a) Find the probability that in a given year there will be less than 21 earthquakes.
This is the pvalue of Z when X = 21. So
has a pvalue of 0.5160.
So there is a 51.60% probability that in a given year there will be less than 21 earthquakes.
b) Find the probability that in a given year there will be between 18 and 23 earthquakes.
This is the pvalue of Z when X = 23 subtracted by the pvalue of Z when X = 18. So:
X = 23
has a pvalue of 0.7611
X = 18
has a pvalue of 0.2676
So there is a 0.7611 - 0.2676 = 0.4935 = 49.35% probability that in a given year there will be between 18 and 23 earthquakes.