Tags are attached to the left and right hind legs of a cow in a pasture. Let A1 be the event that the left tag is lost and the e
1 answer:
Answer:
Step-by-step explanation:
Given that tags are attached to the left and right hind legs of a cow in a pasture. Let A1 be the event that the left tag is lost and the event that the right leg tag is lost. Suppose those two events are independent and A2 the event that the right leg tag is lost. Suppose those two events are independent and P(A1) = P(A2) = 0.3
a) the probability that at least one leg tag is lost
=
(since A1 and A2 are independent)
=0.3+0.3-0.09
=0.51
b) the probability that exactly one tag is lost, given that at least one tag is lost.
= P(one leg lost)/P(atleast one leg lost)
=
c) the probability that exactly one tag is lost, given that at most one tag is lost.
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Answer:
0.1587
Step-by-step explanation:
Let X be the commuting time for the student. We know that . Then, the normal probability density function for the random variable X is given by
. We are seeking the probability P(X>35) because the student leaves home at 8:25 A.M., we want to know the probability that the student will arrive at the college campus later than 9 A.M. and between 8:25 A.M. and 9 A.M. there are 35 minutes of difference. So,
= 0.1587
To find this probability you can use either a table from a book or a programming language. We have used the R statistical programming language an the instruction pnorm(35, mean = 30, sd = 5, lower.tail = F)