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Records kept in a hospital show that 3/10 casualties who come to the casualty department have to wait more than 1/2 hr before receiving medical attention. Find, to 3 d.p., the probability that of the first 8 casualties who come to that casualty department: a) none b) more than 2 will have to wait more than 1/2 hr before receiving medical attention. Find also the most probable number of the 8 casualties that will have to wait mroe than half an hour.

a) was ok, I got the correct answer but b) I did X ~ B(8, 0.3) P(X>>2) got 0.745 and I ended up 4 being most probable, but 0.745 doesn't match the answer at the back - the answer says 0.448.

a) was ok, I got the correct answer but b) I did X ~ B(8, 0.3) P(X>>2) got 0.745 and I ended up 4 being most probable, but 0.745 doesn't match the answer at the back - the answer says 0.448.

You misread "at least 2" for "more than 2" and so worked out P(X >= 2) instead of P(X > 2).

The most likely number of long-waiting casualties is the number x that maximizes P(X = x):

P(X = 0) = 0.0576480,

P(X = 1) = 0.197650,

P(X = 2) = 0.296475 (most likely),

P(X = 3) = 0.254122,

P(X = 4) = 0.136137,

P(X = 5) = 0.0466754,

P(X = 6) = 0.0100019,

P(X = 7) = 0.00122472,

P(X = 8) = 0.00006561.

That's a slow way of getting the answer. Another way is to guess that the most likely number of long-waiting casualties x will be near the mean of X, which is 8*0.3 = 2.4. So we guess x = 2 since that is the integer closest to 2.4. Then we calculate

P(X = 1) = 0.197650,

P(X = 2) = 0.296475,

P(X = 3) = 0.254122.

Since the middle probability is bigger than the other two, it follows that our guess was right.

The most likely number of long-waiting casualties is the number x that maximizes P(X = x):

P(X = 0) = 0.0576480,

P(X = 1) = 0.197650,

P(X = 2) = 0.296475 (most likely),

P(X = 3) = 0.254122,

P(X = 4) = 0.136137,

P(X = 5) = 0.0466754,

P(X = 6) = 0.0100019,

P(X = 7) = 0.00122472,

P(X = 8) = 0.00006561.

That's a slow way of getting the answer. Another way is to guess that the most likely number of long-waiting casualties x will be near the mean of X, which is 8*0.3 = 2.4. So we guess x = 2 since that is the integer closest to 2.4. Then we calculate

P(X = 1) = 0.197650,

P(X = 2) = 0.296475,

P(X = 3) = 0.254122.

Since the middle probability is bigger than the other two, it follows that our guess was right.

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can someone please explain what principle domain is and why the answer is a not c?Maths

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