Answer:
the probability that one parachute deploys is P(one parachute deploys)= 0.9988 (99.88%)
Step-by-step explanation:
for event A= first parachute deploys , then P(A)=0.97
for event B= second parachute deploys, then
P(C∩B)= P(B/C)*P(C)
where
event C= first parachute fails
P(C) = 1-P(A)= 1-0.97 =0.03
P(B/C) = probability that second parachute deploys, when the fist one fails =0.96
P(C∩B)= probability that fist parachute fails and second parachute deploys
then
P(C∩B)= P(B/C)*P(C)=0.03*0.96 = 0.0288
then
P(one parachute deploys)= P(A) + P(C∩B) - P(A∩(C∩B)) ,
but since A and C are mutually exclusive events P(A∩(C∩B)) =0
thus
P(one parachute deploys)= P(A) + P(C∩B) =0.97 + 0.0288 = 0.9988
P(one parachute deploys)= 0.9988
