An oil prospector will drill a succession of holes in a given area to find a productive well. The probability that he is successful on a given trial is 0.2.

Question

a) What's the probability that that the third hole drilled is the first to yield a productive well?
b) If the prospector can afford to drill at most 10 wells, what's the probability that he will fail to find a productive well?

2)Refer to Prob. 1. The prospector drills holes until he finds a productive well. How many holes would the prospector expect to drill? Interpret your answer intuitively.

My answers:
1. a) P(x=3)=.128 (used geom. dist)
b)0.1074(used binomial dist.)

2. I really don't even know how to start this one.

Answer 1

a) so you want
F,F,S
prob = (.8)(.8)(.2) = .128 ---- you were correct

b) you want 10 consecutive failures
prob = .8^10 = .1073... ------ you are correct again

2. This question falls under the catergory:
Number of trials until we have success

Here is a page that details it, and develops the formula
Number of trials to success = 1/p

http://www.cut-the-knot.org/Probability/LengthToFirstSuccess.shtml

in our case p = .2
so the number of trials needed to expect success = 1/.2 = 5

Answer 2

Show that the curvature is greatest on this curve at the same point at which the absolute value of the torsion is greatest. Find that point, and evaluate both the curvature and the torsion there. Also, find the three principal unit vectors there: tangent, normal, and binormal. Note that the torsion is equal to -1/2sech(t)=-curvature K(represents curvature)
we know that the absolute value of the torsion is greatest when K is greatest
Also, we know that since torsion=-curvature then torsion is always negative which means that B always turns towards N rather than -N