Answer:
The probability of failure of both radios is 0.00003364.
The flight becomes much safer with two independent radios
Step-by-step explanation:
When you have two events A and P with probabilities P(A)=q and P(B)=r, if the events are independent (that is to say, if the occurrence of one event does not influence the occurrence of the other), then
[tex]P(A\cap B)=P(A)P(B)[/tex]
where the event intersection means the probability of the occurrence of both events.
If we call P(A) the probability of a radio failure, as the occurrence of failure of a radio is independent of having another radio, then the probability of the failure of both radios is
(0.0058)*(0.0058) = 0.00003364
So, the probability that a particular flight will be threatened with the failure of both radios is 0.00003364.
Hence the flight becomes much safer with two independent radios