A food safety laboratory tested 200 samples of organic and conventional produce to detect … — SAT

A food safety laboratory tested $200$ samples of organic and conventional produce to detect the presence of a specific agricultural residue. The table below summarizes the results.

Produce Type	Residue Detected	Residue Not Detected	Total
Organic	$15$	$65$	$80$
Conventional	$45$	$75$	$120$
Total	$60$	$140$	$200$

If one of the tested samples is selected at random, given that the sample had no residue detected, what is the probability that the selected sample is conventional produce?

A
$\frac{3}{8}$
B
$\frac{5}{8}$
$\frac{15}{28}$ Cevap
D
$\frac{13}{28}$

Cevap

The correct probability is

\frac{15}{28}

The correct answer is

\frac{15}{28}

. To find the conditional probability that a randomly selected sample is conventional produce given that it has no residue detected, we restrict our focus to the column labeled 'Residue Not Detected'. The total number of samples in this column is

140

(

65

organic +

75

conventional). Out of these

140

samples,

75

are conventional produce. The probability is therefore

\frac{75}{140}

, which simplifies to

\frac{15}{28}

by dividing both the numerator and the denominator by

5

Adım Adım Çözüm

Identify the total number of outcomes that satisfy the given condition.

The total number of samples with no residue detected is

65 + 75 = 140

Since the question asks for the probability 'given that the sample had no residue detected', the sample space is restricted to only the column 'Residue Not Detected'.

Identify the number of favorable outcomes within the restricted sample space.

Within the group of samples with no residue detected, the number of conventional produce samples is

75

We need to find the count where the sample is conventional produce under the condition that no residue was detected.

Calculate the conditional probability and simplify the fraction.

\text{Probability} = \frac{75}{140} = \frac{15}{28}

The probability is the ratio of favorable outcomes to the total outcomes in the restricted sample space. Dividing the numerator and denominator by their greatest common divisor,

5

, yields

\frac{15}{28}

Anahtar Kavram

Conditional Probability and Two-Way Tables

Tahmini Süre:1m 15s

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Cevap

Adım Adım Çözüm

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