probability question that should be simple?
A drag racer has 2 paracutes, a main and a backup, that are designed to bring the verhicle to a stop after the end of a run. Suppose that the main chute deploys with the probability of .99 and 5that if the main fails to deploy the backup deploys with a probability of 0.98
a) what is the probability that one of the two parachutes deploys?
b) what is the probability that the backup parachute deploys?
can someone please show me how to do these?
Givens:
P(main deploys) = P(M) = 0.99
∴ P(main does not deploy) = P(not M) = 0.01
As an independent event,
P(backup deploys) = P(BU) = 0.98
P(backup does not deploy) = P(not BU) = 0.02
let us look at b) first: what is the probability that the backup parachute deploys?
However, the second event, the "backup deploys" event, is dependent on the Main deploying occurring first. So taken together as two dependent events:
IF Main fails, and backup deploys:
P(main fails and backup deploys)
= P(not M) x P(BU)
= 0.01 x 0.98
= 0.0098
∴ P(the backup parachute deploys) = 0.98% <1%
now for a) next:
what is the probability that one of the two parachutes deploys?
Two methods:
Method A:
ie P(hat one of the two parachutes deploys)?
There are 3 cases: both deploy, none deploy, or one or the other deploys
P(both deploy) = P(M and BU)
= 0.99 x 0.98
= 0.9702
P(neither deploy) = P(notM and notBU)
= 0.01 x 0.02
=0.0002
so what's left of certainty [ ie P(one of these three cases will occur) = 100% = 1] is :
P(one ot the other chute deploys)
= 1 - 0.9702 - 0.0002
= 0.0296
= 2.96%
Method B:
= P(main opens) x P(not BU) + P(not M) x P(BU)
= 0.99 x 0.02 + 0.01 x 0.98
= 0.0198 + 0.0098
= 0.0296
the probability that one of the two parachutes deploys is 2.96%
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