Question

U and V are mutually exclusive events. P(U) = 0.26; P(V) = 0.37. Find:

a. P(U AND V)

b. P(U|V)

c. P(U OR V)

Short answer

a. $P(UANDV)=0$

b. $P\left(U\right|V)=0$

c.$P(UORV)=0.63$

Step-by-step solution

Content Introduction

Two events are said to be mutually exclusive in probability theory if they cannot occur at the same time or concurrently. In other words, discontinuous events are events that are mutually exclusive. If two occurrences are considered discontinuous, the chances of both happening at the same time are nil.

Explanation(Part a)

We are given, $P\left(U\right)=0.26andP\left(V\right)=0.37$

It is given that U and V are mutually exclusive. Therefore,

$P(\hspace{0.17em}UAND\hspace{0.17em}V)=0$

Explanation (Part b)

Here, $P\left(U\right|V)$can be calculates as follow:

$$\begin{gathered} P\left(U\right|V)=\frac{P(UAND\hspace{0.17em}V)}{P\left(V\right)} \\ P(U\left|V\right)=\frac{0}{0.37} \\ P\left(U\right|V)=0 \end{gathered}$$

Explanation(Part c)

It is given that U and V are mutually exclusive. Therefore,

$$\begin{gathered} P(U\hspace{0.17em}OR\hspace{0.17em}V)=P\left(U\right)+P\left(V\right) \\ P(U\hspace{0.17em}OR\hspace{0.17em}V)=0.26+0.37 \\ P(U\hspace{0.17em}OR\hspace{0.17em}V)=0.63 \end{gathered}$$