This Exam P sample reference tests Conditional Probability. This problem conditions two independent discrete counts on their combined total. The favorable weight is 0.0064 out of a total weight of 0.0424, giving 8/53=0.150943 and choice C.
How to solve this Conditional Probability question
Setup
Setup
Let X and Y denote the counts in the two periods. Independence makes the probability of any ordered pair equal to the product of its two marginal probabilities.
Pr(X=i,Y=j)=Pr(X=i)Pr(Y=j)
Model
Model
A combined count of two can arise through exactly three ordered pairs. The middle pair is the favorable event.
{X+Y=2}={(0,2),(1,1),(2,0)}
Pr(X=1,Y=1)=0.082=0.0064
Compute
Compute
Add the three disjoint weights and divide the favorable weight by their sum.
Pr(X+Y=2)=0.90(0.02)+0.082+0.02(0.90)=0.0424
Pr(X=1,Y=1∣X+Y=2)=0.04240.0064=538=0.1509433962
Answer
Answer
The conditional probability rounds to 0.151.
0.151(C)
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These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThe value 0.006 is the rounded joint probability 0.08²=0.0064 before conditioning; it omits division by the probability of the observed total.
BThe value 0.042 is the rounded denominator 0.0424, the probability of a combined count of two, rather than the requested conditional ratio.
DUsing only (0,2) and (1,1) in the denominator gives 0.0064/[0.90(0.02)+0.0064]=0.2623. This omits the distinct ordered outcome (2,0).
EThe calculation 1-2(0.02)=0.960 merely removes the two single-period double-count probabilities from one; it neither restricts the sample space to a total of two nor forms a conditional ratio.
Original practice · fully worked
Original variant: access digits under a mismatch condition
An authentication tester independently generates a first digit uniformly from 1, 2, 3, 4, and 5 and a second digit uniformly from 1, 2, 3, and 4. Given that the two generated digits are different, calculate the probability that their sum is 5.
A 1/5
B 1/4
C 4/15
D 4/9
E 3/4
Variant answer in brief
There are 20 equally likely ordered digit pairs, and removing the four equal pairs leaves 16 under the condition. Four of those pairs sum to 5, so the conditional probability is 1/4 and choice B.
Setup
Setup
Represent the generated digits by an ordered pair. The unequal ranges produce 5 × 4 equally likely pairs.
#{(X,Y)}=5(4)=20
Model
Model
The condition removes the four possible equal pairs; a pair with two fives is unavailable because the second digit cannot be five.
#{X=Y}=20−4=16
Compute
Compute
Enumerate the sum-five pairs. Every one of them also satisfies the condition.
The 3108-page Probability Proof Manual reorganizes 718 verified Exam P solutions by syllabus skill and adds formula proofs, error patterns, and original worked practice.
