This Exam P sample reference tests Mixture Probability. Conditioning on the selected die gives squared six-probabilities 1/36, 1/36, 1/9, and 1. Their equally weighted average is 7/24=0.292, choice C.
Condition on which die is selected. Given a die, the two rolls are independent with that die's stated six-probability.
p1=p2=61,p3=31,p4=1
Model
Model
For each die, square its one-roll six-probability to obtain the conditional probability of two sixes. Then average the four conditional probabilities because the die is selected uniformly.
Pr(two sixes)=41j=1∑4pj2
Compute
Compute
The four conditional probabilities sum to seven sixths. Dividing by four gives seven twenty-fourths, or approximately 0.292.
Pr(two sixes)=41(361+361+91+1)=247
Answer
Answer
The probability of two sixes is about 0.292.
0.292(C)
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BThe value 0.250 is only the contribution of selecting the die that always rolls a six, before adding the other three dice's positive contributions.
EThe value 0.417 is the average probability of a six on one roll. It does not square the die-specific probabilities for two rolls.
Original practice · fully worked
Original variant: repeated alerts from one selected sensor
One of three sensors is selected uniformly and used for two independent trials. The sensors have per-trial alert probabilities 0.20, 0.50, and 0.90. Find the probability that both trials produce an alert.
A 0.3000
B 0.3333
C 0.3667
D 0.5333
E 0.6000
Variant answer in brief
Conditioning on the sensor gives an average of squared alert probabilities: (0.04+0.25+0.81)/3=0.3667, choice C.
Setup
Setup
Condition on the selected sensor. Once a sensor is selected, the two trials are independent with the same sensor-specific alert probability.
p=(0.20,0.50,0.90)
Model
Model
Square each sensor's alert probability to obtain its conditional two-alert probability, then average because the three sensors are equally likely.
Pr(two alerts)=31(0.202+0.502+0.902)
Compute
Compute
The squared probabilities sum to 1.10. Dividing by three gives approximately 0.3667.
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