Assume 75% of the AP stats students studied for this test. If 40% of those who studied get an A but only 10% of those who did not study get A, what is the probability that someone who gets an A actually studied for the test?

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Assume 75% of the AP stats students studied for this test. If 40% of those who studied get an A but only 10% of those who did not study get A, what is the probability that someone who gets an A actually studied for the test?

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Answer 1 · 2018-05-21T12:00:47

Let $A =$ $A=$ the student get an A in test
Let $B =$ $B=$ the student studied for test

We know that $P (B) = 0.75$ $P(B)=0.75$ and $P (A / B) = 0.40$ $P(A//B)=0.40$

We know also that $P (A / B) = \frac{P (A \cap B)}{P (B)} = 0.40$ $P(A//B)=(P(AnnB))/(P(B))=0.40$

We are looking for $P (A \cap B) = 0.40 \times 0.75 = 0.3$ $P(AnnB)=0.40xx0.75=0.3$

Answer 2 · 2018-05-26T15:45:27

The probability is (approximately) 92.31%.

Explanation:

Let $A$ $A$ be the event that the student got an A on the test.
Let $B$ $B$ be the event that the student studied.

It sounds like we are seeking the conditional probability of, "studied for the test GIVEN THAT they got an A".

This is written as $P (B ∣ A)$ $"P"(B | A)$ , and it is read as "probability of B given A."

We are told:

$P (B) = 0.75$ $"P"(B) = 0.75$ which means $P (B^{C}) = 0.25$ $"P"(B^"C") = 0.25$
$P (A ∣ B) = 0.4$ $"P"(A|B)=0.4$
$P (A ∣ B^{C}) = 0.1$ $"P"(A|B^"C") = 0.1$
(the $^{C}$ $""^"C"$ means "complement")

and we want to find $P (B ∣ A)$ $"P"(B|A)$ . By definition,

$P (B ∣ A) = \frac{P (B \cap A)}{P (A)}$ $"P"(B|A) = ("P"(B nn A))/("P"(A))$

Using Bayes' Theorem, we can rewrite this as

$P (B ∣ A) = \frac{P (A ∣ B) P (B)}{P (A ∣ B) P (B) + P (A ∣ B^{C}) P (B^{C})}$ $"P"(B|A) = ("P"(A|B)"P"(B))/("P"(A|B) "P"(B)+"P"(A|B^"C") "P"(B^"C"))$

Every probability on the right side is now something we know. We plug these values in:

$P (B ∣ A) = \frac{0.4 \times 0.75}{0.4 \times 0.75 + 0.1 \times 0.25}$ $"P"(B|A) = (0.4 xx 0.75)/(0.4 xx 0.75" " +" "0.1 xx 0.25)$

$P (B ∣ A) = \frac{0.3}{0.3 + 0.025}$ $color(white)("P"(B|A)) = (0.3)/(0.3+0.025)$

$P (B ∣ A) = \frac{0.3}{0.325}$ $color(white)("P"(B|A)) = (0.3)/(0.325)$

$P (B ∣ A) = \frac{12}{13} \approx 0.9231$ $color(white)("P"(B|A)) = 12/13" " ~~ 0.9231$

So there is a 92.31% chance that, given that a student got an A, they also studied.

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Assume 75% of the AP stats students studied for this test. If 40% of those who studied get an A but only 10% of those who did not study get A, what is the probability that someone who gets an A actually studied for the test?

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