Let [tex]\( E \)[/tex] and [tex]\( F \)[/tex] be two events in an experiment with the sample space [tex]\( S \)[/tex]. We are given that [tex]\( P(E) = 0.5 \)[/tex], [tex]\( P(F) = 0.4 \)[/tex], and [tex]\( P(E \cap F) = 0.1 \)[/tex]. We need to compute the following probabilities step-by-step:
### (a) [tex]\( P(E \cup F) \)[/tex]
The probability of the union of two events [tex]\( E \)[/tex] and [tex]\( F \)[/tex] is given by:
[tex]\[ P(E \cup F) = P(E) + P(F) - P(E \cap F) \][/tex]
Substituting in the given values:
[tex]\[ P(E \cup F) = 0.5 + 0.4 - 0.1 \][/tex]
So:
[tex]\[ P(E \cup F) = 0.8 \][/tex]
### (b) [tex]\( P(E^C) \)[/tex]
The complement of [tex]\( E \)[/tex], denoted as [tex]\( E^C \)[/tex], is the event that [tex]\( E \)[/tex] does not occur. The probability of the complement of [tex]\( E \)[/tex] is given by:
[tex]\[ P(E^C) = 1 - P(E) \][/tex]
Substituting in the given value of [tex]\( P(E) \)[/tex]:
[tex]\[ P(E^C) = 1 - 0.5 \][/tex]
So:
[tex]\[ P(E^C) = 0.5 \][/tex]
### (c) [tex]\( P(F^C) \)[/tex]
The complement of [tex]\( F \)[/tex], denoted as [tex]\( F^C \)[/tex], is the event that [tex]\( F \)[/tex] does not occur. The probability of the complement of [tex]\( F \)[/tex] is given by:
[tex]\[ P(F^C) = 1 - P(F) \][/tex]
Substituting in the given value of [tex]\( P(F) \)[/tex]:
[tex]\[ P(F^C) = 1 - 0.4 \][/tex]
So:
[tex]\[ P(F^C) = 0.6 \][/tex]
### (d) [tex]\( P(E^C \cap F) \)[/tex]
To find the probability of [tex]\( E^C \cap F \)[/tex], the event that [tex]\( E \)[/tex] does not occur but [tex]\( F \)[/tex] does occur, we use:
[tex]\[ P(E^C \cap F) = P(F) - P(E \cap F) \][/tex]
Substituting in the given values:
[tex]\[ P(E^C \cap F) = 0.4 - 0.1 \][/tex]
So:
[tex]\[ P(E^C \cap F) = 0.3 \][/tex]
Finally, the computed probabilities are:
- [tex]\( P(E \cup F) = 0.8 \)[/tex]
- [tex]\( P(E^C) = 0.5 \)[/tex]
- [tex]\( P(F^C) = 0.6 \)[/tex]
- [tex]\( P(E^C \cap F) = 0.3 \)[/tex]
