This Exam P sample reference tests Independence. Writing the two-trial exactly-one probability as x=2p(1-p), the four-trial exactly-two probability becomes 6p²(1-p)²=(3/2)x². Thus choice D is correct.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThe target must be quadratic in pq and hence quadratic in x. For example, p=1/2 gives x=1/2 and the true probability 3/8, whereas 0.5x gives 1/4.
BThis linearly doubles the two-period probability. At p=1/2 it gives 2x=1, while the exactly-two count has probability 6/16=3/8.
CRequiring exactly one success in each fixed two-period block gives x²=4p²q². It misses the two arrangements in which one block has two successes and the other has none.
EThe value 3x²=12p²q² doubles all six valid arrangements. Each binary sequence with two successes already has probability p²q² and must be counted once.
Original practice · fully worked
Original variant: conditional online-sensor count
Five sensors operate independently during a diagnostic window. For each sensor, the odds of being online to being offline are 2:3. The diagnostic reports that at least one sensor is online. Calculate the conditional probability that exactly two sensors are online.
A 0.07776
B 0.34560
C 0.37474
D 0.68256
E 0.92224
Variant answer in brief
The odds give an online probability of 0.4. Dividing the binomial exactly-two mass 0.34560 by the nonzero-count probability 0.92224 gives 0.37474, so choice C is correct.
Setup
Setup
Convert the individual odds into online and offline probabilities.
p=2+32=0.4,q=53=0.6
Model
Model
Let K be the online count. Independence gives a binomial model, then the diagnostic conditions on K being positive.
K∼Binomial(5,0.4)
Pr(K=2∣K≥1)=Pr(K≥1)Pr(K=2)
Compute
Compute
Calculate the numerator and the conditioning probability separately.
Pr(K=2)=(25)(0.4)2(0.6)3=0.34560
Pr(K≥1)=1−(0.6)5=0.92224
Pr(K=2∣K≥1)=0.922240.34560=0.3747397641…
Answer
Answer
The requested conditional probability rounds to 0.37474.
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