Conditional probability and Bayes Test

{"questions":[{"content":"Given $P(C\\cap D)=0.12$ and $P(D)=0.3$, enter $P(C\\mid D)$ as a decimal. [[input-conditional-probability-and-bayes-test-1]]","widgets":{"input-conditional-probability-and-bayes-test-1":{"type":"input","answer":"0.4"}},"hints":[],"explanation":"The conditional probability is 0.12/0.3.","id":"conditional-probability-and-bayes-test-1","evidence_kind":"objective-concept-test","correct_feedback":"The conditional probability is 0.12/0.3.","incorrect_feedback":"D is the condition, so divide the intersection by P(D), not P(C). This measures the fraction of D that also belongs to C."},{"content":"In a group, 24 study music, of whom 9 also study art. Enter the probability of art given music as a decimal. [[input-conditional-probability-and-bayes-test-2]]","widgets":{"input-conditional-probability-and-bayes-test-2":{"type":"input","answer":"0.375"}},"hints":[],"explanation":"Use 9/24, because the condition selects the music students.","id":"conditional-probability-and-bayes-test-2","evidence_kind":"objective-concept-test","correct_feedback":"Use 9/24, because the condition selects the music students.","incorrect_feedback":"The condition selects the 24 music students. Only the 9 who also study art are successes within that restricted group."},{"content":"A and B have probabilities 0.6 and 0.5 and intersection probability 0.3. Are they independent?\u003Cbr\u003E[[choice-conditional-probability-and-bayes-test-3]]","widgets":{"choice-conditional-probability-and-bayes-test-3":{"type":"choice","options":["Yes","No","Only if they are disjoint"],"answer":[0],"choice_feedback":["The intersection equals 0.6(0.5).","Compare P(A and B) with P(A)P(B): both equal 0.3. Overlap does not prevent independence.","Positive-probability independent events are generally not disjoint. Here their intersection equals the required product."]}},"hints":[],"explanation":"The intersection equals 0.6(0.5).","id":"conditional-probability-and-bayes-test-3","evidence_kind":"objective-concept-test","correct_feedback":"The intersection equals 0.6(0.5)."},{"content":"A supplier provides 40% of components with defect rate 0.05; another provides 60% with rate 0.02. Enter the overall defect probability as a decimal. [[input-conditional-probability-and-bayes-test-4]]","widgets":{"input-conditional-probability-and-bayes-test-4":{"type":"input","answer":"0.032"}},"hints":[],"explanation":"Total probability is 0.4(0.05) + 0.6(0.02).","id":"conditional-probability-and-bayes-test-4","evidence_kind":"objective-concept-test","correct_feedback":"Total probability is 0.4(0.05) + 0.6(0.02).","incorrect_feedback":"Weight each defect rate by its supplier share before adding. An unweighted average would incorrectly assume equal supplier shares."},{"content":"Using supplier shares 40% and 60% and defect rates 0.05 and 0.02, enter the probability a defective component came from the first supplier as a decimal. [[input-conditional-probability-and-bayes-test-5]]","widgets":{"input-conditional-probability-and-bayes-test-5":{"type":"input","answer":"0.625"}},"hints":[],"explanation":"The first supplier's defect path is 0.02; divide by total defect probability 0.032.","id":"conditional-probability-and-bayes-test-5","evidence_kind":"objective-concept-test","correct_feedback":"The first supplier's defect path is 0.02; divide by total defect probability 0.032.","feedback_visual":{"type":"probability_tree","priors":[0.4,0.6],"rates":[0.05,0.02],"labels":["Supplier 1","Supplier 2"],"outcome":"defect","caption":"The defect paths are 0.02 and 0.012. Once a defect is known, these two paths make the new total 0.032; the first supplier contributes 0.02/0.032."},"incorrect_feedback":"The numerator is the first supplier's share times its defect rate. Divide by defects from both suppliers, not by the first supplier's share alone."},{"content":"An alert detects 90% of faulty items. Why is it invalid to say 90% of alerted items are faulty?\u003Cbr\u003E[[choice-conditional-probability-and-bayes-test-6]]","widgets":{"choice-conditional-probability-and-bayes-test-6":{"type":"choice","options":["The fault prevalence and false-alert rate are also needed","Detection rates cannot be probabilities","The probability must instead be 10%"],"answer":[0],"choice_feedback":["The reverse conditional depends on both the base rate and alerts from non-faulty items.","Detection rates are conditional probabilities and are valid probabilities. The missing information concerns the reverse conditional.","Taking a complement changes positive to negative within the same condition. It does not reverse the condition from faulty to alerted."]}},"hints":[],"explanation":"The reverse conditional depends on both the base rate and alerts from non-faulty items.","id":"conditional-probability-and-bayes-test-6","evidence_kind":"objective-concept-test","correct_feedback":"The reverse conditional depends on both the base rate and alerts from non-faulty items.","deeper_explanation":"An alert can come from either a faulty or a non-faulty item. The 90% detection rate describes only the faulty branch. To reverse the conditional, both branch contributions and their starting proportions are needed."},{"content":"Box A is chosen with probability p and gives red with probability 0.8. Box B gives red with probability 0.2. If $P(\\text{red})=0.5$, enter p as a decimal. [[input-conditional-probability-and-bayes-test-7]]","widgets":{"input-conditional-probability-and-bayes-test-7":{"type":"input","answer":"0.5"}},"hints":[],"explanation":"Solve 0.8p + 0.2(1 − p) = 0.5.","id":"conditional-probability-and-bayes-test-7","evidence_kind":"objective-concept-test","correct_feedback":"Solve 0.8p + 0.2(1 − p) = 0.5.","incorrect_feedback":"Use the total-probability equation 0.8p + 0.2(1 − p) = 0.5. The second source has probability 1 − p, not p."},{"content":"A source is faulty with probability 0.1. Two checks are conditionally independent given its status; each is positive with probability 0.8 if faulty and 0.2 otherwise. Given two positives, find the probability it is faulty.\u003Cbr\u003E[[choice-conditional-probability-and-bayes-test-8]]","widgets":{"choice-conditional-probability-and-bayes-test-8":{"type":"choice","options":["$16/25$","$4/13$","$9/10$"],"answer":[0],"choice_feedback":["The two paths have probabilities 0.1(0.8²) = 0.064 and 0.9(0.2²) = 0.036. Normalize to 0.64.","4/13 would use only one positive check. With two conditionally independent checks, multiply the two conditional positive rates within each status branch.","0.9 is the prior probability of being non-faulty. After two positives, compare the faulty and non-faulty two-positive paths instead."]}},"hints":[],"explanation":"The two paths have probabilities 0.1(0.8²) = 0.064 and 0.9(0.2²) = 0.036. Normalize to 0.64.","id":"conditional-probability-and-bayes-test-8","evidence_kind":"objective-concept-test","correct_feedback":"The two paths have probabilities 0.1(0.8²) = 0.064 and 0.9(0.2²) = 0.036. Normalize to 0.64.","deeper_explanation":"Conditional independence lets you multiply the two positive rates within each status: $0.8^2$ if faulty and $0.2^2$ otherwise. Weight by the status probabilities, then normalize $0.064/(0.064+0.036)=0.64$.","feedback_visual":{"type":"probability_tree","priors":[0.1,0.9],"rates":[0.64,0.04],"labels":["Faulty","Not faulty"],"outcome":"two positives","caption":"The second-stage rates already combine the two conditionally independent checks: 0.8² and 0.2². The faulty/two-positive path is 0.064; the other is 0.036."}}],"mix":0,"tinkumo_module_mode":"test"}