This Exam P sample reference tests Conditional Probability. The beta-type survival function is (1-y)⁵. Dividing the tails at 0.4 and 0.1 gives (2/3)⁵=0.13169, choice B.
How to solve this Conditional Probability question
Setup
Setup
Normalize the beta-type density first. Integrating over the unit interval gives normalizing constant five.
1=k∫01(1−y)4dy,k=5
Model
Model
Integrating the density above a threshold gives the fifth power of one minus that threshold. Because the event above 0.4 is contained in the event above 0.1, the conditional probability is the ratio of those two survival values.
S(y)=∫y15(1−t)4dt=(1−y)5
Compute
Compute
The numerator and denominator are the fifth powers of 0.6 and 0.9. Their ratio is approximately 0.131687.
Pr(Y>0.4∣Y>0.1)=0.950.65=0.131687
Answer
Answer
The conditional probability is approximately 0.1317, which rounds to 0.13 and selects choice B.
0.13(B)
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AThe value 0.08 is the unconditional probability of exceeding 0.4, rounded from 0.07776. It omits division by the probability of exceeding 0.1.
Original practice · fully worked
Original variant: conditional median above a utilization threshold
A utilization fraction Y has density f(y)=3(1-y)² for 0<y<1. Given that utilization exceeds 0.20, find the conditional median m, meaning P(Y>m | Y>0.20)=0.50.
A 0.3000
B 0.3650
C 0.4000
D 0.5000
E 0.6000
Variant answer in brief
The survival function is (1-y)³. Solving the conditional tail equation gives m=1-0.8×2⁻¹⁄³=0.3650, choice B.
Setup
Setup
Integrating the density above a threshold gives the cube of one minus that threshold.
S(y)=∫y13(1−t)2dt=(1−y)3
Model
Model
The conditional median makes the survival probability above the median exactly half the survival probability above 0.20.
S(0.20)S(m)=0.83(1−m)3=0.50
Compute
Compute
Solving the resulting cube equation gives a median of approximately 0.365040.
m=1−0.8(0.50)1/3=0.365040
Answer
Answer
The conditional median is approximately 0.3650, selecting choice B.
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