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.
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.
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
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
Well, well, well, let's drill into these questions, shall we?
1. a) To find the probability that the third hole drilled is the first to yield a productive well, we can use the geometric distribution. The probability of success on any given trial is 0.2, so the probability of failure is 0.8. The geometric distribution formula is P(x=k) = (1-p)^(k-1)*p, where k is the number of trials until the first success. In this case, we want to find P(x=3), so plug in the values: P(3) = (1-0.2)^(3-1)*0.2. Simplifying this gives us P(3) = 0.128.
b) To calculate the probability that the prospector fails to find a productive well in 10 attempts, we can use the binomial distribution. The formula is P(x=k) = nCk * p^k * (1-p)^(n-k), where n is the number of trials, k is the number of successes, and p is the probability of success on any given trial. In this case, n = 10, k = 0 (since the prospector fails to find a productive well), and p = 0.2. Plugging in these values, we get P(0) = 10C0 * 0.2^0 * (1-0.2)^(10-0) = 0.1074.
2. Ah, the expected number of holes to be drilled. This can be calculated using the expected value formula for a geometric distribution. The formula is E(x) = 1/p, where p is the probability of success on any given trial. In this case, p = 0.2, so the expected number of holes to be drilled is E(x) = 1/0.2 = 5.
Intuitively, this means that the prospector can expect to drill around 5 holes before finding a productive well on average. But hey, don't take my word for it, drilling is a slippery business!
To find the expected number of holes the prospector would drill until finding a productive well, we can use the concept of expected value in probability theory. The expected value is also known as the mean or average value.
In this case, let's denote the random variable X as the number of holes drilled until finding a productive well. We know that the probability of finding a productive well on each trial is 0.2, so the probability of not finding a productive well on each trial is 0.8.
To calculate the expected number of holes, we can use the formula:
E(X) = 1/p,
where p is the probability of success on a single trial.
In this case, p = 0.2. Therefore, the expected number of holes is:
E(X) = 1/0.2 = 5.
So, the prospector would expect to drill approximately 5 holes until finding a productive well.
Intuitively, this means that on average, the prospector would need to drill 5 holes before finding a productive well. This is because the probability of finding a productive well on each trial is 0.2, so it takes several trials on average to find a successful one.
To solve the given problems, you correctly used the geometric distribution for part a) and the binomial distribution for part b). Now, let's move on to problem 2).
Problem 2)
The prospector keeps drilling holes until he finds a productive well. We are asked to find the expected number of holes the prospector would need to drill.
To solve this problem, we can use the concept of expected value. The expected value represents the average value of a random variable.
In this case, we can define the random variable X as the number of holes the prospector needs to drill until finding a productive well.
Since the prospector has a 0.2 probability of finding a productive well on each drilling attempt, the probability distribution of X can be represented by the geometric distribution with a success probability of 0.2.
The probability mass function (PMF) of the geometric distribution is given by:
P(X = k) = (1 - p)^(k-1) * p
Where p is the probability of success (0.2) and k represents the number of trials needed to achieve the first success.
Now, to find the expected value of X (denoted as E[X]), we can use the following formula:
E[X] = 1/p
In this case, the expected value of X would be:
E[X] = 1/0.2 = 5
Hence, the prospector would be expected to drill an average of 5 holes until finding a productive well.
Intuitively, this means that on average, the prospector would need to drill 5 holes before striking oil. It gives us an estimate of the number of drilling attempts required before achieving success.