marinacalder
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Hi , I've been struggling with these 3 questions for a while now, could someone please offer a solution?
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marinacalder
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(Original post by marinacalder)
Hi , I've been struggling with these 3 questions for a while now, could someone please offer a solution?

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marinacalder
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(Original post by marinacalder)
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old_engineer
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For Q16 I would suggest starting by constructing separate probability distribution tables for X and Y. The rest should then fall into place.
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Physics Enemy
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(Original post by marinacalder)
...
16) p(h) = p(t) = 0.5, p(d = 1, 2, … 6) = 1/6
i) P(h = 5, X wins) = 1 - [P(d = 5) + P(d = 6)]
ii) P(d = 3, X wins) = P(h = 4) + P(h = 5) + P(h = 6)
Note: P(h = r) = 6Cr (0.5)^r (0.5)^(6 - r) = 6Cr 0.5^6

17) p(c) = 0.9, p(f) = 0.1; independent students
i) 15 completions in a row => (0.9)^15
ii) 15C1 ways of 1 fail, each of p = (0.1)(0.9)^14
iii) 15C2 ways of 2 fails, each of p = (0.2)^2(0.9)^13
P(Max: f = 2) = P(f = 0) + P(f = 1) + P(f = 2)
iv) P(Min: f = 2) = 1 - [P(f = 0) + P(f = 1)]

18) p(h) = 4/5, p(m) = 1/5; independent shots
i) P(Min: h = 8) = P(h = 8) + P(h = 9) + P(h = 10)
Note: P(h = r) = 10Cr (4/5)^r (1/5)^(10 - r)
ii) P(Max: h = 7) = 1 - P(Min: h = 8)
iii) 10C7 ways of 7 hits & 3 misses
Can miss 1-3, 2-4, ... 8-10: 8 ways
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