P(E) = 0.38
P(F) = 0.57
The of this question is to find the probability of two mutually exclusive events E and F when either of them can occur.
The question is based on the concept of probability of mutually exclusive events. Two events are mutually exclusive events when both of these events do not occur at the same time, for example when a die is rolled or when we toss a coin. The probability that it will come head or tail is completely separate from each other. These two events can not occur at the same time, it will be either head or tail. These types of events are called mutually exclusive events.
Expert Answer
The probability that either E or F will occur can be calculated by adding the probabilities of both of the events. The probabilities of the separate events is given as:
\[ P (E) = 0.38 \]
\[P (F) = 0.57 \]
The probability of two mutually exclusive events occurring at the same time is given by:
\[ P( E\ and\ F) = 0 \]
As these two events are mutually exclusive, their probability of occurring at the same time is always zero.
The probability that either of these mutually exclusive events will occur is given by:
\[ P ( E\ or\ F ) = P (E) + P (F) \]
\[ P ( E\ or\ F ) = 0.38 + 0.57 \]
\[ P ( E\ or\ F ) = 0.95 \]
The probability that either E or F will occur is 0.95 or 95%.
Numerical Result
The probability that either two mutually exclusive events E and F will occur is calculated to be:
\[ P ( E\ or\ F ) = 0.95 \]
Example
Find the probability P ( G or H ), if G and H are two mutually exclusive events. The probabilities of the separate events are given below:
\[ P (G) = 0.43 \]
\[ P (H) = 0.41 \]
The probability that either G or H will occur can be calculated by adding the probabilities of both of the events.
The probability that either of these mutually exclusive events will occur is given by:
\[ P ( G\ or\ H ) = P (E) + P (F) \]
\[ P ( G\ or\ H ) = 0.43 + 0.41 \]
\[ P ( G\ or\ H ) = 0.84 \]
The probability of G and H, two mutually exclusive events, when either of these events can occur is calculated to be 0.84 or 84%.