For each exam, Ariadne studies with probability 1/2 and does not study with probability 1/2, independently of any other exams. On any exam for which she has not studied, she still has a 0.20 probability of passing, independently of whatever happens on other exams. What is the expected number of total exams taken until she has had 3 exams for which she did not study but which she still passed?
30
any answer?
Well, let's calculate how many exams it takes for Ariadne to have 3 exams for which she did not study but still passed.
On any individual exam, there are 2 outcomes: she studies and passes, or she doesn't study and still passes. If she studies, the exam is counted as one that she has studied. If she doesn't study, but still passes, the exam is counted as one that she hasn't studied but passed.
The probability of her passing an exam she studied for is 1/2, and the probability of passing an exam she didn't study for is 0.20.
Let's expect the number of exams it takes for her to have 3 exams for which she didn't study but still passed. To get this expectation, we need to calculate the expected value for the number of exams.
Let's define X as the number of exams she has taken until she has had 3 exams for which she did not study but passed.
Let's consider two cases: she either studies for an exam or she doesn't study.
Case 1: She studies for an exam.
In this case, she has taken one exam (X = 1) and we still need to have 2 more exams for which she didn't study but passed. The probability of passing such exams is (0.20)^2.
Case 2: She doesn't study for an exam.
In this case, she has taken one exam (X = 1) and we still need to have 2 more exams for which she didn't study but passed. The probability of passing such exams is (0.20)^2.
Therefore, X = 1 + (0.20)^2 * X
Simplifying the equation, we get X = 1 + 0.04X
Rearranging, we find 0.96X = 1
Dividing both sides by 0.96, we get X = 1.0417
Therefore, the expected number of total exams taken until Ariadne has had 3 exams for which she did not study but still passed is approximately 1.0417 exams.
But hey, who needs to study anyway? Ariadne manages to pass exams without even studying! Maybe she should become a professional "exam-passer"!
Let's break down the problem step-by-step:
Step 1: Determine the probability of passing an exam without studying.
We are given that the probability of passing an exam without studying is 0.20.
Step 2: Determine the probability of not passing an exam without studying.
Since the probability of passing an exam without studying is 0.20, the probability of not passing is 1 - 0.20 = 0.80.
Step 3: Determine the probability of Ariadne not studying for an exam.
We are given that Ariadne studies with a probability of 1/2 and does not study with a probability of 1/2. Therefore, the probability of not studying is 1/2.
Step 4: Calculate the probability of Ariadne not passing an exam for which she did not study.
To find this probability, we multiply the probability of not studying (1/2) by the probability of not passing an exam without studying (0.80). The result is:
Probability of not passing an exam = (1/2) * (0.80) = 0.40
Step 5: Calculate the expected number of exams until Ariadne has had 3 exams for which she did not study but still passed.
Let X be the number of exams until Ariadne has had 3 exams for which she did not study but still passed.
To calculate the expected value, we need to consider two scenarios:
- If Ariadne studies for an exam, the expected number of exams remaining is still X.
- If Ariadne does not study for an exam, the expected number of exams remaining is X - 1.
Since Ariadne decides to study with a probability of 1/2, the probability of each scenario is:
- Probability of studying = 1/2
- Probability of not studying = 1/2
Using the principle of linearity of expectation, the expected number of exams until Ariadne has had 3 exams for which she did not study but still passed can be expressed as:
E(X) = (1/2) * X + (1/2) * (X - 1)
Simplifying the equation:
E(X) = (1/2) * X + (1/2) * X - 1/2
E(X) = X - 1/2
Step 6: Solve for the expected number of exams.
Since we know that Ariadne has to take 3 exams for which she did not study but still passed, we set X - 1/2 equal to 3 and solve for X:
X - 1/2 = 3
X = 3 + 1/2
X = 7/2
Therefore, the expected number of total exams taken until Ariadne has had 3 exams for which she did not study but still passed is 7/2 or 3.5 exams.
To find the expected number of total exams taken until Ariadne has 3 exams for which she did not study but still passed, we can break down the problem into smaller sub-problems.
Let's define a random variable X_i as follows:
X_i = 1 if the i-th exam is one for which Ariadne did not study but still passed
X_i = 0 otherwise
We want to find the expected value of the random variable N, where N represents the total number of exams taken until Ariadne has had 3 exams for which she did not study but passed.
N can be defined as follows:
N = X_1 + X_2 + X_3 + ...
To find the expected value E(N), we can use the linearity of expectation:
E(N) = E(X_1) + E(X_2) + E(X_3) + ...
Now, let's calculate the probabilities and expected values for X_i:
For each exam, Ariadne studies with probability 1/2 and does not study with probability 1/2. Therefore, the probability that she did not study but still passed is 1/2 times 0.20 = 0.10.
So, the probability distribution for X_i is as follows:
P(X_i = 1) = 0.10
P(X_i = 0) = 1 - 0.10 = 0.90
Now, let's calculate the expected value E(X_i) for each i:
E(X_i) = 1 * P(X_i = 1) + 0 * P(X_i = 0)
= 1 * 0.10 + 0 * 0.90
= 0.10
Now, we can substitute the values of E(X_i) into the equation for E(N):
E(N) = E(X_1) + E(X_2) + E(X_3) + ...
= 0.10 + 0.10 + 0.10 + ...
= 0.10 * (1 + 1 + 1 + ...)
= 0.10 * ∞
Since we want the expected number of exams taken until Ariadne has had 3 exams for which she did not study but passed, we know that this process will eventually stop after exactly 3 such exams. Therefore, the expected value E(N) can be calculated as follows:
E(N) = 0.10 * 3
= 0.30
So, the expected number of total exams taken until Ariadne has had 3 exams for which she did not study but passed is 0.30.