Answer:
(a) The value of <em>x</em> is 0.30.
(b) The probability that a reported case of rabies is not a raccoon is 0.45.
(c) The probability that a reported case of rabies is either a bat or a fox is 0.15.
Step-by-step explanation:
Denote the events as follows:
<em>R</em> = reported case of rabies is a raccoon
<em>F</em> = reported case of rabies is a fox
<em>B</em> = reported case of rabies is a bat
<em>O</em> = reported case of rabies is a some other animal.
The data provided is:
P (R) = 0.55
P (F) = 0.11
P (B) = 0.04
P (O) = <em>x</em>.
(a)
A property of a probability distribution is that the sum of all individual properties is 1.
That is, ![\sum P(X)=1](https://tex.z-dn.net/?f=%5Csum%20P%28X%29%3D1)
Compute the value of <em>x</em> as follows:
![\sum P(X)=1\\P(R)+P(F)+P(B)+P(O)=1\\0.55+0.11+0.04+x=1\\0.70+x=1\\x=1-0.70\\x=0.30](https://tex.z-dn.net/?f=%5Csum%20P%28X%29%3D1%5C%5CP%28R%29%2BP%28F%29%2BP%28B%29%2BP%28O%29%3D1%5C%5C0.55%2B0.11%2B0.04%2Bx%3D1%5C%5C0.70%2Bx%3D1%5C%5Cx%3D1-0.70%5C%5Cx%3D0.30)
Thus, the value of <em>x</em> is 0.30.
(b)
The probability of the complement of an event is the probability of its not happening.
![P(A^{c})=1-P(A)](https://tex.z-dn.net/?f=P%28A%5E%7Bc%7D%29%3D1-P%28A%29)
Compute the probability that a reported case of rabies is not a raccoon as follows:
![P(R^{c})=1-P(R)\\=1-0.55\\=0.45](https://tex.z-dn.net/?f=P%28R%5E%7Bc%7D%29%3D1-P%28R%29%5C%5C%3D1-0.55%5C%5C%3D0.45)
Thus, the probability that a reported case of rabies is not a raccoon is 0.45.
(c)
Compute the probability that a reported case of rabies is either a bat or a fox as follows:
![P(B\cup F)=P(B)+P(F)\\=0.11+0.04\\=0.15](https://tex.z-dn.net/?f=P%28B%5Ccup%20F%29%3DP%28B%29%2BP%28F%29%5C%5C%3D0.11%2B0.04%5C%5C%3D0.15)
Thus, the probability that a reported case of rabies is either a bat or a fox is 0.15.