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D 1 is independent of D 2 and D 3 . Pr[D 1 ] = 0.8. Pr[D 2 ] = 0.8. Pr[D 3 |D 2

ID: 3432185 • Letter: D

Question

D1 is independent of D2 and D3.
Pr[D1] = 0.8.
Pr[D2] = 0.8.
Pr[D3|D2] = 0.9.
Pr[D3|Dc2] = 0.5.

(a) Find Pr[D3], the probability that the third dart hits the target.
(b) Given that the third dart hits the target, what is the probability that the second dart
hit the target?
(c) Find the probability that all three darts hit the target.
(d) Find the probability that exactly two darts hit the target.

You throw three darts at a target. Let D_{i} be the event that dart i hits the target. Suppose you are given the following information: (a) Find Pr[D3], the probability that the third dart hits the target. (b) Given that the third dart hits the target, what is the probability that the second dart hit the target? (c) Find the probability that all three darts hit the target. (d) Find the probability that exactly two darts hit the target.

Explanation / Answer

(a) Find Pr[D3], the probability that the third dart hits the target.

Pr[D3|D2] = 0.9

--> P(D3 and D2)/P(D2) =0.9

--> P(D3 and D2)/0.8 = 0.9

--> P(D3 and D2) =0.9*0.8=0.72

Pr[D3|Dc2] = 0.5

--> P(D3 and D2') / P(D2') =0.5

--> P(D3 and D2')/(1-0.8)=0.5

--> P(D3 and D2') = 0.5*0.2=0.1

--> P(D3)- P(D3 and D2) =0.1

So P(D3) = 0.1+0.72 = 0.82

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(b) Given that the third dart hits the target, what is the probability that the second dart
hit the target?

P(D2|D3) = P(D2 and D3)/P(D3)

=0.72/0.82

=0.8780488

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(c) Find the probability that all three darts hit the target.

P(D1 and D2 and D3) = P(D1)*P(D2 and D3) (D1 is independent of D2 and D3)

=0.8*0.72

=0.576

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(d) Find the probability that exactly two darts hit the target.

P(exactly two) = P(D1 and D2)+P(D1 and D3)+P(D2 and D3)

=P(D1)*P(D2) + P(D1)*P(D3)+ P(D2 and D3)

=0.8*0.8+0.8*0.82 + 0.72

=2.016