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25 Sep 2025, 07:15
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P(X only): 0.35
P(Both): 0.4
In a genetic database of 200 samples, each sample may or may not show Marker X, and each sample may or may not show Marker Y. Every sample that does not show Marker X also does not show Marker Y. Exactly 150 samples show at least one marker, and exactly 120 samples do not show Marker Y.
One sample is selected at random. Select for P(X only) the probability that the sample shows Marker X but not Marker Y, and select for P(Both) the probability that the sample shows both markers. Make only two selections, one in each column.
One sample is selected at random. Select for P(X only) the probability that the sample shows Marker X but not Marker Y, and select for P(Both) the probability that the sample shows both markers. Make only two selections, one in each column.
| P(X only) | P(Both) | |
| 0.2 | ||
| 0.3 | ||
| 0.35 | ||
| 0.4 | ||
| 0.6 | ||
| 0.7 |
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Official Answer
P(X only): 0.35
P(Both): 0.4
25 Sep 2025, 07:15
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Official Solution:
The possible states for a sample are:
• (X yes, Y yes)
• (X yes, Y no)
• (X no, Y no)
The state (X no, Y yes) is impossible, because the rule says if X is absent then Y must also be absent.
Total = 200.
Since 120 samples do not show Y (the Y no groups), the remaining 80 show Y. These must be (X yes, Y yes). So, (X yes, Y yes) = 80.
Also, 150 samples show at least one marker, which must be X, because Y cannot appear without X. Therefore, X = 150, which means (X yes, Y yes) + (X yes, Y no) = 150. This makes (X no, Y no) = 200 − 150 = 50.
Now we can break down the counts:
• (X yes, Y yes) = 80
• (X yes, Y no) = 200 − 80 − 50 = 70
• (X no, Y no) = 50
P(X only) = 70/200 = 0.35
P(Both) = 80/200 = 0.40
Answer: P(X only) = 0.35, P(Both) = 0.40
Correct answer:
P(X only) "0.35"
P(Both) "0.4"
Bunuel
The possible states for a sample are:
• (X yes, Y yes)
• (X yes, Y no)
• (X no, Y no)
The state (X no, Y yes) is impossible, because the rule says if X is absent then Y must also be absent.
Total = 200.
Since 120 samples do not show Y (the Y no groups), the remaining 80 show Y. These must be (X yes, Y yes). So, (X yes, Y yes) = 80.
Also, 150 samples show at least one marker, which must be X, because Y cannot appear without X. Therefore, X = 150, which means (X yes, Y yes) + (X yes, Y no) = 150. This makes (X no, Y no) = 200 − 150 = 50.
Now we can break down the counts:
• (X yes, Y yes) = 80
• (X yes, Y no) = 200 − 80 − 50 = 70
• (X no, Y no) = 50
P(X only) = 70/200 = 0.35
P(Both) = 80/200 = 0.40
Answer: P(X only) = 0.35, P(Both) = 0.40
Correct answer:
P(X only) "0.35"
P(Both) "0.4"
01 Oct 2025, 06:10
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Let's solve using venn diagram
Exactly 150 samples show at least one marker, it means that null value = 50
exactly 120 samples do not show Marker Y, it means only(X)=70
P(X only) = 70/200 = 0.35
P(Both) = 0.4
GMAT-Club-Forum-xj8vwjg2.png [ 6.23 KiB | Viewed 542 times ]
Exactly 150 samples show at least one marker, it means that null value = 50
exactly 120 samples do not show Marker Y, it means only(X)=70
P(X only) = 70/200 = 0.35
P(Both) = 0.4
Bunuel
Attachment:
GMAT-Club-Forum-xj8vwjg2.png [ 6.23 KiB | Viewed 542 times ]
