20 points 4. A new security system needs to be evaluated in the airport. The pro
ID: 3305640 • Letter: 2
Question
20 points 4. A new security system needs to be evaluated in the airport. The probability of a person being a security hazard is 4%. At the checkpoint, the security system denied person without security problems 2% of the time. But the security system passed a person with security problems 1% of the time. a. What is the probability that a person passes through the system? b. What is the probability that a person who passes through the system is without any 15 points 5. You have one box with 4 white balls and 2 red balls, and you randomly keep picking balls without replacement. If X is a random variable representing the number of trials required for the 1st red ball, what is the probability mass function of X?Explanation / Answer
14. Let A denote a person being security hazard. Let B denote a person passing through the system
Given P(A) = 0.04 P(Bc|Ac) = 0.02 P(B|A) = 0.01
P(Ac) = 1 - 0.04 = 0.96 P(B|Ac) = 1 - 0.02 = 0.98
a. P(A B) = P(B|A) P(A) = 0.01 * 0.04 = 0.0004
P(Ac B) = P(B|Ac) P(Ac) = 0.98 * 0.96 = 0.9408
P(B) = P(A B) + P(Ac B) = 0.0004 + 0.9408 = 0.9412
b. P(Ac|B) = P(Ac B) / P(B) = 0.9408 / 0.9412 = 0.9996
15. In general, P(x) will be probability of white balls picked in the first x - 1 trials and red ball picked in the xth trial.
f(1) = 2/6 = 2 * 4P0 / 6P1
f(2) = 4/6 * 2/5 = 2 * 4P1 / 6P2
f(3) = 4/6 * 3/5 * 2/4 = 2 * 4P2 / 6P3
f(4) = 4/6 * 3/5 * 2/4 * 2/3 = 2 * 4P3 / 6P4
f(5) = 4/6 * 3/5 * 2/4 * 1/3 * 1 = 2 * 4P4 / 6P5
The probability mass function is 2 * 4P(x-1) * 6Px
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