test yourself! practice tool\nmrs. dalys reading classes were surveyed about their reading preferences. the…

test yourself! practice tool\nmrs. dalys reading classes were surveyed about their reading preferences. the data is summarized in the table below.\n| | prefers fiction novels | prefers non - fiction novels | total |\n| ---- | ---- | ---- | ---- |\n| 9th grade students | 35 | 8 | 43 |\n| 10th grade students | 52 | 12 | 64 |\n| total | 87 | 20 | 107 |\ngiven that a student prefers fiction novels, what is the probability that they are also in 10th grade based on this data? write your answer as a decimal rounded to the nearest hundredth.\ncalculator

test yourself! practice tool\nmrs. dalys reading classes were surveyed about their reading preferences. the data is summarized in the table below.\n| | prefers fiction novels | prefers non - fiction novels | total |\n| ---- | ---- | ---- | ---- |\n| 9th grade students | 35 | 8 | 43 |\n| 10th grade students | 52 | 12 | 64 |\n| total | 87 | 20 | 107 |\ngiven that a student prefers fiction novels, what is the probability that they are also in 10th grade based on this data? write your answer as a decimal rounded to the nearest hundredth.\ncalculator

Answer

Explanation:

Step1: Recall conditional - probability formula

The formula for conditional probability is $P(A|B)=\frac{P(A\cap B)}{P(B)}$. In the context of this problem, let $A$ be the event that a student is in 10th grade and $B$ be the event that a student prefers fiction novels. Then $P(A|B)=\frac{n(A\cap B)}{n(B)}$, where $n(A\cap B)$ is the number of students who are in 10th grade and prefer fiction novels, and $n(B)$ is the number of students who prefer fiction novels.

Step2: Identify values from the table

From the table, $n(A\cap B) = 52$ (the number of 10th - grade students who prefer fiction novels) and $n(B)=87$ (the total number of students who prefer fiction novels).

Step3: Calculate the probability

$P(A|B)=\frac{52}{87}\approx0.5977$. Rounding to the nearest hundredth, we get $0.60$.

Answer:

$0.60$