This Exam P sample reference tests Discrete Random Variables. Let N be the given count. The target event within the condition is 1≤N≤3, whose probability is 0.045, while the conditioning event N≤3 has probability 0.965. Their ratio is 9/193=0.046632, which matches choice A.
How to solve this Discrete Random Variables question
Setup
Setup
Translate the two verbal events into bounds on the count N. At least one affected unit means N is positive, while at least one unaffected unit rules out the maximum count of four.
{at least one affected}={N≥1}
{at least one unaffected}={N≤3}
Model
Model
Intersect the target with the conditioning event and apply the conditional-probability ratio.
Pr(N≥1∣N≤3)=Pr(N≤3)Pr(1≤N≤3)
Compute
Compute
Add the appropriate entries from the supplied probability distribution.
Pr(1≤N≤3)=0.015+0.010+0.020=0.045
Pr(N≤3)=0.920+0.015+0.010+0.020=0.965
0.9650.045=1939=0.0466321244…
Answer
Answer
The conditional probability rounds to 0.0466.
0.0466(A)
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BThe value 0.0800 is the unconditional probability P(N≥1)=1-P(N=0), so it fails to impose the condition that N cannot equal four.
CThe value 0.0829 divides 0.080 by 0.965 but leaves the excluded N=4 mass in the numerator after removing it from the conditioning denominator.
DThe value 0.5625 equals 0.045/0.080 and reverses the conditioning, calculating the chance of at least one unaffected unit given at least one affected unit.
EThe value 0.7500 equals 0.015/0.020 and arises by replacing both 'at least one' statements with exactly-one point probabilities before taking an unrelated ratio.
Original practice · fully worked
Original variant: delayed shipment count
Let N be the number of delayed shipments in a five-shipment dispatch. Its distribution is P(N=0)=0.64, P(N=1)=0.16, P(N=2)=0.09, P(N=3)=0.06, P(N=4)=0.03, and P(N=5)=0.02. Given that at least one shipment arrives on time, calculate the probability that N is odd.
A 0.220000
B 0.224490
C 0.240000
D 0.244898
E 0.775510
Variant answer in brief
The condition removes N=5 and has probability 0.98. Within that event, an odd delayed count means N=1 or N=3, with total mass 0.22; dividing gives 0.224490 and choice B.
Setup
Setup
At least one on-time shipment means not all five shipments are delayed.
H={N≤4}
Pr(H)=1−Pr(N=5)=1−0.02=0.98
Model
Model
Inside the conditioning event, the possible odd values are one and three; the odd value five has been excluded.
{N odd}∩H={N=1}∪{N=3}
Compute
Compute
Add the two eligible point probabilities and normalize by the conditioning probability.
Pr(N odd,H)=0.16+0.06=0.22
Pr(N odd∣H)=0.980.22=0.2244897959…
Answer
Answer
The conditional probability of an odd delayed count is approximately 0.224490.
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