Respuesta :
a) If the student misses his alarm, the probability that he is not late to school is 0.25.
b) The probability that the student is late to school is 0.65.
c) One day, the student shows up to school on time. The probability that he missed his alarm is 0.59.
For A,
Let A be the event that the student wakes up on time, and B be the event that the student is late to school. We want to find P(B).
By definition,
P(B) = 1 - P(A)
where A is the event that the student misses his alarm.
Therefore,
P(B) = 1 - P(A)
= 1 -0.75
We are given that P(A') = 0.75, so
P(B) =0.25
For B,
To calculate the probability that the student is late to school, we will use the following formula:
P(X = x) = n!/(x!(n-x)!) * p^x * (1-p)^(n-x)
where:
P(X = x) is the probability that the student is late to school
n is the number of trials
x is the number of events
p is the probability of an event occurring
For this problem, we will let n = 1 and x = 1.
Now that we have our formula, we can plug in our values to solve for P(X = x).
P(X = x) = 1!/(1!(1-1)!) * 0.65^1 * (1-0.65)^(1-1)
P(X = x) = 0.65
We can interpret this answer as follows: there is a 0.65 probability that the student is late to school.
For C,
We first need to find the probability that the student did not miss his alarm, which is 1 - 0.59 = 0.41. We then need to find the probability that the student showed up to school on time given that he did not miss his alarm, which is 1 - 0.41 = 0.59.
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