Answer: .11
Step-by-step explanation:
Let F be the event that the first truck is available and S be the event that the second truck is available.
The probability of neither truck being available is expressed as P([tex]F^{C}[/tex]∩[tex]S^{C}[/tex]) , where P([tex]F^{C}[/tex]) is the probability that the event F doesn't happen and P([tex]S^{C}[/tex]) is the probability that the event S doesn't happen.
P([tex]F^{C}[/tex])= 1-P(F) = 1-0.73 = 0.27
P([tex]S^{C}[/tex])=1-P(S) = 1-0.59 = 0.41
Since [tex]F^{C}[/tex] and [tex]S^{C}[/tex] aren't mutually exclusive events, then:
P([tex]F^{C}[/tex]∪[tex]S^{C}[/tex]) = P([tex]F^{C}[/tex]) + P([tex]S^{C}[/tex]) - P([tex]F^{C}[/tex]∩[tex]S^{C}[/tex])
Isolating the probability that interests us:
P([tex]F^{C}[/tex]∩[tex]S^{C}[/tex])= P([tex]F^{C}[/tex]) + P([tex]S^{C}[/tex])- P([tex]F^{C}[/tex]∪[tex]S^{C}[/tex])
Where P([tex]F^{C}[/tex]∪[tex]S^{C}[/tex]) = 1 - 0.43 = 0.57
Finally:
P([tex]F^{C}[/tex]∩[tex]S^{C}[/tex]) = 0.27+ 0.41 - 0.57 = 0.11