Answer:
6.57, 6.38, 6.125, 6.057, 6.03, 6.01
Answer:
(a) 0.9412
(b) 0.9996 ≈ 1
Step-by-step explanation:
Denote the events a follows:
= a person passes the security system
= a person is a security hazard
Given:
Then,
(a)
Compute the probability that a person passes the security system using the total probability rule as follows:
The total probability rule states that: 
The value of P (P) is:
Thus, the probability that a person passes the security system is 0.9412.
(b)
Compute the probability that a person who passes through the system is without any security problems as follows:
Thus, the probability that a person who passes through the system is without any security problems is approximately 1.
Using binomial distribution where success is the appearing of any of the top 10 most common names, thus probability of success (p) is 9.6% = 0.096 and the probability of failure = 1 - 0.096 = 0.904. Number of trials is 11.
Binomial distribution probability is given by P(x) = nCx (p)^x (q)^(n - x)
Probability that none of the top 10 most common names appears is P(0) = 11C0 (0.096)^0 (0.904)^(11 - 0) = (0.904)^11 = 0.3295
Thus, the probability that at least one of the 10 most common names appear is 1 - 0.3295 = 0.6705
Therefore, I will be supprised that none of the names of the authors were among the 10 most common names given that the probability that at least one of the names appear is 67%.
If $.72 per 1lb, then $x per 2.7lb, by using proportional property:
.72/1 = x/2.7 cross multiply
x = (.72)*(2.7) = $1.94
