Answer To: 3.14 Two fair dice are tossed, and the up face on each die is MI recorded. a. List the 36 sample...
David answered on Dec 22 2021
3.14)
a.) The 36 sample points are given below:
(1,1),(1,2),(1,3),(1,4),(1,5),(1,6)
(2,1),(2,2),(2,3),(2,4),(2,5),(2,6)
(3,1),(3,2),(3,3),(3,4),(3,5),(3,6)
(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)
(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)
(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)
b.) Note that, for any pair (x,y) {x=1,2,3,4,5,6 and y=1,2,3,4,5,6}
P(x appears)=1/6 and P(y appears)=1/6
Hence P((x,y) appears)=P(x appears) P(y appears) [ since they are independent]
=1/36
Thus we probability of each of the pairs occurring is 1/36.
c.)
(A) The only favorable case is (3,3) whose probability is 1/36.
(B) The favorable cases are
(1,1),(1,3),(1,5),(2,2),(2,4),(2,6),(3,1),(3,3),(3,5),(4,2),(4,4),(4,6),(5,1),(5,3),(5,5),(6,2),(6,4),(6,6)
Each has probability 1/36 and hence total probability is the sum of them which is 18/36=1/2.
(C) The favorable cases are
(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)
Each has probability 1/36 and hence total probability is the sum of them which is 6/36=1/6.
(D) The favorable cases are
(1,5),(2,5),(3,5),(4,5),(5,1), (5,2),(5,3),(5,4),(5,5),(5,6),(6,5)
Each has probability 1/36 and hence total probability is the sum of them which is 11/36.
(E) The favorable cases are
(4,6),(5,5),(5,6),(6,4),(6,5),(6,6)
Each has probability 1/36 and hence total probability is the sum of them which is 6/36=1/6.
3.21)
a.) Out of 188 beech trees surveyed, 49 had been damaged by fungi.
Hence the probability of a beech tree in East Central Europe being damaged by fungi
=49/188=0.26
b.) The Probability that trunk of the selected tree would be damaged by fungi =
...