QUESTION IMAGE
Question
based on the percentages in the tree diagram, is the new test a reliable indicator that a child with a positive test actually has a food allergy? use the drop-down menus to explain.
actually has allergy?
yes 8%
no 92%
test shows allergy?
for yes: yes 90%, no 10%
for no: yes 15%, no 85%
click the arrows to choose an answer from each menu.
the probability that a child will receive a positive test is choose... . for a child with a positive test, the probability that the child actually has a food allergy is approximately choose... . the new test choose... likely to be a reliable indicator that the child has a food allergy.
Step1: Calculate P(Positive Test)
To find the probability of a positive test, we consider both cases where the child actually has the allergy and where they don't.
For the "Yes" (has allergy) branch: The probability of having the allergy is \( 8\% = 0.08 \), and the probability of testing positive given they have the allergy is \( 90\% = 0.90 \). So the joint probability here is \( 0.08 \times 0.90 = 0.072 \).
For the "No" (doesn't have allergy) branch: The probability of not having the allergy is \( 92\% = 0.92 \), and the probability of testing positive given they don't have the allergy is \( 15\% = 0.15 \). So the joint probability here is \( 0.92 \times 0.15 = 0.138 \).
Now, we sum these two joint probabilities to get the total probability of a positive test: \( 0.072 + 0.138 = 0.21 \) or \( 21\% \).
Step2: Calculate P(Has Allergy | Positive Test)
We use Bayes' theorem here. The formula for conditional probability is \( P(A|B) = \frac{P(A \cap B)}{P(B)} \). Here, \( A \) is "has allergy" and \( B \) is "positive test". We already found \( P(A \cap B) = 0.072 \) (from Step 1) and \( P(B) = 0.21 \) (from Step 1). So we calculate \( \frac{0.072}{0.21} \approx 0.3429 \) or approximately \( 34.3\% \) (or we can keep it as a fraction: \( \frac{0.072}{0.21}=\frac{72}{210}=\frac{12}{35}\approx 0.34 \)).
Step3: Determine Reliability
Since the probability that a child with a positive test actually has the allergy is around \( 34\% \), which is less than \( 50\% \) (or not very high), the test is not very reliable.
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The probability that a child will receive a positive test is \( 21\% \) (or \( 0.21 \)). For a child with a positive test, the probability that the child actually has a food allergy is approximately \( 34\% \) (or \( \frac{12}{35} \approx 0.34 \)). The new test is not likely to be a reliable indicator that the child has a food allergy.
(If we need to fill in the drop - down menus as per the problem's context, the first "Choose..." would be "21%", the second "Choose..." would be "34%" (or the approximate value), and the third "Choose..." would be "not".)