Over time, a student analyzes her ability to guess correctly after narrowing down multiple choice answers in a 5-selection question (a through e). She discovers that if she narrows her answers to 2 or 3 choices, her probability of getting the right answer is 0.8, but if she still has 4 or 5 choices left, her probability of choosing correctly decreases drastically to 0.1. Assuming that in general she can narrow his choices to 2-3 choices 70% of the time, answer the following probability questions.

a. What is the probability that she will answer a question correctly any time she must guess?

b. Given that the student answered correctly, what is the probability that she had narrowed her choices to 2-3?

20%

59% - Verified

To answer these questions, we need to use conditional probability. Let's denote the events as follows:

A = Correctly answering a question
B = Narrowing choices to 2-3

a. What is the probability that she will answer a question correctly any time she must guess?

We can use the Law of Total Probability to find the probability of answering correctly regardless of narrowing choices.
P(A) = P(A|B)P(B) + P(A|B')P(B')

Given the information, we have:
P(A|B) = 0.8 (probability of answering correctly given narrowing choices to 2-3)
P(B) = 0.7 (probability of narrowing choices to 2-3)

P(A|B') = 0.1 (probability of answering correctly given not narrowing choices to 2-3)
P(B') = 1 - P(B) = 1 - 0.7 = 0.3 (probability of not narrowing choices to 2-3)

Substituting the values:
P(A) = (0.8)(0.7) + (0.1)(0.3)
= 0.56 + 0.03
= 0.59

Therefore, the probability that she will answer a question correctly any time she must guess is 0.59 or 59%.

b. Given that the student answered correctly, what is the probability that she had narrowed her choices to 2-3?

We can use Bayes' theorem to find the conditional probability of narrowing choices to 2-3 given that she answered correctly.
P(B|A) = (P(A|B)P(B)) / P(A)

Using the values we have:
P(A|B) = 0.8 (probability of answering correctly given narrowing choices to 2-3)
P(B) = 0.7 (probability of narrowing choices to 2-3)
P(A) = 0.59 (probability of answering correctly regardless of narrowing choices)

Substituting the values:
P(B|A) = (0.8)(0.7) / 0.59
= 0.56 / 0.59
≈ 0.9492

Therefore, given that the student answered correctly, the probability that she had narrowed her choices to 2-3 is approximately 0.9492 or 94.92%.

To solve these probability questions, we can use the concept of conditional probability. Let's break down the information given:

Let A be the event that the student narrows her choices to 2-3.
Let B be the event that the student answers a question correctly.

a. We want to find the probability that the student answers a question correctly, which is P(B).

We are given:
P(B | A) = 0.8 (the probability of answering correctly given she narrows choices to 2-3)
P(B | A') = 0.1 (the probability of answering correctly given she doesn't narrow choices to 2-3)
P(A) = 0.7 (the probability of narrowing choices to 2-3)

We can use the Law of Total Probability to calculate P(B):

P(B) = P(B | A) * P(A) + P(B | A') * P(A')

P(B) = 0.8 * 0.7 + 0.1 * (1 - 0.7)
P(B) = 0.56 + 0.03
P(B) = 0.59

Therefore, the probability that the student answers a question correctly when she must guess is 0.59.

b. We want to find the probability that the student had narrowed her choices to 2-3, given that she answered correctly, which is P(A | B).

We can use Bayes' theorem to calculate P(A | B):

P(A | B) = (P(B | A) * P(A)) / P(B)

We already know:
P(B | A) = 0.8
P(B) = 0.59
P(A) = 0.7

P(A | B) = (0.8 * 0.7) / 0.59
P(A | B) ≈ 0.9441

Therefore, given that the student answered correctly, the probability that she had narrowed her choices to 2-3 is approximately 0.9441.