This Exam P sample reference tests Conditional Distributions. This is a conditional-expectation calculation from a joint probability table. Restricting the table to the qualifying rows gives probability 0.70 and a weighted numerator of 0.99, so the conditional mean is 99/70, or 1.414286, and choice C is correct.
How to solve this Conditional Distributions question
Setup
Setup
Retain the two rows allowed by the conditioning event. Their combined probability is the denominator of the conditional distribution.
Pr(X<2)=0.30+0.40=0.70
Model
Model
For each possible value of Y, add the retained joint masses and divide by 0.70. The conditional mean can be computed by dividing the corresponding weighted joint sum by the same denominator.
E[Y∣X<2]=Pr(X<2)∑yyPr(Y=y,X<2)
Compute
Compute
The retained column masses for Y equal 0.14, 0.21, 0.27, and 0.08. Weighting them by 0, 1, 2, and 3 produces the numerator.
y∑yPr(Y=y,X<2)=1(0.21)+2(0.27)+3(0.08)=0.99
E[Y∣X<2]=0.700.99=7099=1.414285714…
Answer
Answer
The conditional expected count rounds to 1.41.
1.41(C)
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These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AUsing a multiplier of two rather than three for the Y=3 column gives numerator 0.91 and 0.91/0.70=1.30.
BMisadding the retained Y=3 column as 0.07 instead of 0.08 gives numerator 0.96 and 0.96/0.70=1.3714.
DThis is the unconditional mean of Y from all four rows. It ignores the condition on X and therefore uses the wrong distribution.
EUsing a multiplier of four rather than three for the Y=3 column gives numerator 1.07 and 1.07/0.70=1.5286.
Original practice · fully worked
Original variant: microfluidic completion time
Within a normalized one-minute microfluidic cycle, U records when a sample enters a channel and V records when processing finishes. Their joint density equals 2 on the triangular region 0<U<V<1 and is zero elsewhere. For samples that enter after time 1/2, calculate the expected completion time V.
A 0.2083
B 0.6667
C 0.7500
D 0.8333
E 1.0000
Variant answer in brief
The event U>1/2 has probability 1/4, while the corresponding first-moment integral for V is 5/24. Their ratio is 5/6, or 0.8333, so choice D is correct.
Setup
Setup
The conditioning event restricts U to the interval from one-half to one. For each such entry time, V ranges from U to one.
A={U>1/2}
fU,V(u,v)=2,21<u<v<1
Model
Model
A conditional expectation is the first-moment integral over the restricted region divided by the probability of that region.
E[V∣A]=∫1/21∫u12dvdu∫1/21∫u12vdvdu
Compute
Compute
Evaluate the restricted probability and first moment separately, then normalize.
Pr(A)=∫1/212(1−u)du=41
E[V1A]=∫1/21(1−u2)du=245
E[V∣A]=1/45/24=65=0.833333…
Answer
Answer
The expected normalized completion time is approximately 0.8333.
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