Question

A drag racer has two parachutes, a main and a backup, that are designed to bring the vehicle to a stop after the end of a run. Suppose that the main chute deploys with probability 0.97, and that if the main fails to deploy, the backup deploys with probability 0.96.
1. What is the probability that one of the two parachutes deploys?

Answer 1

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

Answer 2

In this exercise we have to use the knowledge of probability to calculate the chances of having two parachutes delivered, in this way we find that:

99.88%

Simplifying the information given in the statement, we find that:

• for event A= first parachute deploys , then P(A)=0.97

• for event B= second parachute deploys
• event C= first parachute fails

Then using the probability formula equal to:

$P(C\in B)= P(B/C)*P(C)\\ P(C) = 1-P(A)= 1-0.97 =0.03$

Where :

• 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

$P(C B)= P(B/C)*P(C)=0.03*0.96 = 0.0288\\ P(A) + P(CB) =0.97 + 0.0288 = 0.9988$

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