Question

Multi-part question:

Please show explanation and step by step work.

Answer 1

a)

Let A be event that main parachute will be deployed

Let B be event that back up parachute will be deployed

It is given that

P(A)=0.90

As we know that sum of probabilities will be equal to one

P(A)+P(A^c)=1

This implies that P(A^c)=1-P(A)=1-0.90=0.10

P(A^c)=0.10

P(B/A^c)=0.90

We have to find the probability that backup parachute deployed

We will use the law of total probability

P(B)=P(B/A).P(A)+P(B/A^c)P(A^c)

If the main parachute will be deployed the probability that back up parachute deployes equal to zero

that P(B/A)=0

thus Substitute all the values in above law of probability

P(B)=(0)x0.90+0.90x0.10=0+0.09=0.09

So probability that backup parachute deployed will be 0.09

b)

We have to find probability that one of two parachutes will be deployed

The event that one of parachutes will be deployed is the complement of event the none of parachute will be deployed

So we have to find the P(AUB)

$P(A\cup B)=1-P(A^{c}\cap B^{c})$

$=1-P(A^{c})P(B^{c}/A^{}c)$

Now using the multiplication rule

$=1-[1-P(A)][1-P(B/A^{c})]$

=1-(1-0.90)(1-0.90)

=1-(0.10)(0.10)

=1-0.01

=0.99

probability that one of two parachutes will be deployed will be 0.99

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