Probability question. Watch

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Report Thread starter 12 years ago
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"On my way home from work I pass through three sets of traffic lights. The probabilities that I pass through them without having to stop are 0.2, 0.4 and 0.7 respectively. The lights operate independently of each other".

Let S represent "stop" and G represent "go".

i) P(Stop just once) = P(S,G,G) + P(G, S, G) + P(G, G, S)
= 0.224 + 0.034 + 0.024
= 0.332

ii) Given that I have to stop at just one set of lights, find the probability that I have to stop at the first set of lights.

P(Stop at first set of lights | Stop only once) = P(Stop at first lights only)/P(Stop only once)

= P(S,G,G)/0.332
= 0.224/0.332
= 0.675

iii) The probability of passing through the first set of lights without having to stop is changed from 0.2 to p. This changes the answer from part (ii) to 0.5. Calculate p.

So now I know that:

P(Stop at first lights only)/P(Stop only once) = 0.5
P(Stop at first lights only) = 0.5P(Stop only once)
0.224 = 0.5P(Stop only once)
0.448 = P(Stop only once)
0.448 = P(S, G, G) + P(G, S, G) + P(G, G, S).
0.448 = (0.8x0.4x0.7) + (px0.6x0.7) + (px0.4x0.3)
0.448 = 0.224 + 0.42p + 0.12p
0.224 = 0.54p
p = 0.415

The answer in the back of the book says 0.341, what've I done wrong?
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Chewwy
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#2
Report 12 years ago
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"0.448 = (0.8x0.4x0.7) + (px0.6x0.7) + (px0.4x0.3)"

0.8 shouldn't be here
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#3
Report Thread starter 12 years ago
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Ugh...

Thanks chewwy

I'll have to rep you tommorow, I've reached my limit for today apparently.
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#4
Report Thread starter 12 years ago
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I changed it to:

0.448 = {(1-p)x0.4x0.7} + (px0.6x0.7) + (px0.4x0.3)
0.448 = 0.28 - 0.28p + 0.42p + 0.12p
0.168 = 0.26p
p = 0.646 ?

What's gone wrong this time?
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#5
Report Thread starter 12 years ago
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Anybody ?
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