[SOLVED] CS代考计算机代写 Text Problem 4.48

Text Problem 4.48
In some military courts, 9 judges are appointed. However, both the prosecution and the defense attorneys are entitled to a peremptory challenge of any judge, in which case that judge is removed from the case and is not replaced. A defendant is declared guilty if the majority of judges cast votes of guilty, and he or she is declared innocent otherwise. Suppose that when the defendant is, in fact, guilty, each judge will (independently) vote guilty with probability .7, whereas when the defendant is, in fact, innocent, this probability drops to .3.
a. What is the probability that a guilty defendant is declared guilty when there are (i) 9, (ii) 8, and (iii) 7 judges?
b. Repeat part (a) for an innocent defendant.
c. If the prosecuting attorney does not exercise the right to a peremptory challenge of a judge, and if the defense is limited to at most two such challenges, how many challenges should the defense attorney make if he or she is 60 percent certain that the client is guilty?
To solve something like this you pretty much have to fall back on a decent calculator or computer for computation.
If there are 9 judges, and the defendant is guilty, the defendant is declared guilty if 5 or more of the judges cast guilty votes. The number of judges casting guilty votes is a Binomial random variable with n = 9 trials and p = 0.7 probability. So we need to compute
P(X ≥ 5) = p(5) + p(6) + p(7) + p(8) + p(9)
For instance, p (5) =  9  0.750.34 = 0.1715332 . The software package R will compute this:
> dbinom(5,9,0.7)
[1] 0.1715322
[> is a prompt for user input. Ignore the [1].)
You have 4 more probabilities to obtain (then sum).
> dbinom(6,9,0.7)
[1] 0.2668279
> dbinom(7,9,0.7)
[1] 0.2668279
> dbinom(8,9,0.7)
[1] 0.1556496
> dbinom(9,9,0.7)
[1] 0.04035361
P(X ≥ 5) = 0.1715322 + 0.2668279 + 0.2668279 + 0.1556496 + 0.04035361 = 0.9011912.
5 

Instead taking complements:
P(X ≥ 5) = 1 – P(X < 5) = 1 – P(X ≤ 4) = 1 – F(4)where F is the cumulative distribution function. F(4) = p(0)+p(1)+p(2)+p(3)+p(4)= 0.000019683 + 0.000413343 + 0.003857868+ 0.02100395 + 0.07351382 = 0.09880866 SoP(X≥5)=1–F(4) =1–0.09880866 = 0.9011914But it’s much quicker to use Rs built in cumulative distribution function. > pbinom(4,9,0.7)
[1] 0.09880866
Similarly, the (conditional) probability an innocent person is found guilty is
P(X≥5)=1–F(4) =1–0.9011913 = 0.09880866
Note: These two Fs are different functions. One takes p = 0.7; the other p = 0.3. Technically they should be named with different letters – like F0.7 and F0.3.
Here are the guilty probabilities for the given situations (jury size and defendant status). Defendant Status
> dbinom(0,9,0.7)
[1] 1.9683e-05
> dbinom(1,9,0.7)
[1] 0.000413343
> dbinom(2,9,0.7)
[1] 0.003857868
> dbinom(3,9,0.7)
[1] 0.02100395
> dbinom(4,9,0.7)
[1] 0.07351382
Jury size 9
8
7
Guilty
> 1 – pbinom(4,9,0.7)
[1] 0.9011913
> 1 – pbinom(4,8,0.7)
[1] 0.8058956
> 1 – pbinom(3,7,0.7)
[1] 0.873964
Innocent
> 1 – pbinom(4,9,0.3)
[1] 0.09880866
> 1 – pbinom(4,8,0.3)
[1] 0.05796765
> 1 – pbinom(3,7,0.3)
[1] 0.126036
The Guilty column of this table solves part a; the Innocent column part b. To solve part c use the law of total probability.
> pbinom(4,9,0.3)
[1] 0.9011913
> 1 – pbinom(4,9,0.3)
[1] 0.09880866