4. Rumpelstiltskin is spinning straw into gold thread to save the miller\'s daug

Question

4. Rumpelstiltskin is spinning straw into gold thread to save the miller's daughter, for which, according to the Grimm's tale, she must promise to give him her first born son. (Incidentally, this poor woman was never given a name in the story, so we shall call her Beatrix.) Suppose that wise Beatrix made her promise contingent on the quality of his work, and that she knows from an unnamed source that his spun gold thread has on average 2 flaws per meter. a) If the agreement is that Rumple must produce 3 meters of flawless gold thread or he forfeits the child, what is the probability that Beatrix keeps her child? b) If you were Rumple, what is the maximum continuous length of flawless thread you would agree to produce and still have at least a 50% chance of keeping Beatrix's child?

Explanation / Answer

a)expected numebr of flaw in 3 meters =2*3 =6

P(beatrix keeps her child)=P(Rumple made at leasy one flaw) =1-P(0 flaw in 3 meter)

=1-e-660/0! =1-0.0025 =0.9975

b)

let length of flawless thread is x:

threfore P(beatrix keep the child) =1-P(X=0) >0.5

1-e-2x >0.5

e-2x <0.5

taking log and solving:

x >ln(0.5)/2

x>0.347 meter

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