Solution to a. List the sample space in the given experiment. How many outcomes are possible? b. … - Sikademy
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a. List the sample space in the given experiment. How many outcomes are possible? b. Construct a table showing the number of defective computers in each outcome and assign this number to this outcome. What is the value of the random variable X? c. Illustrate a probability distribution. What is the probability value P(X) to each value of the random variable? d. What is the sum of the probabilities of all values of the random variable? e. What do you notice about the probability of each value of the random variable?

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a.There are 2^3=8 possible outcomes


S=\{DDD, DDN, DND, NDD,

DNN,NDN, NND, NNN\}


b. The possible values of the random variable X are 0, 1, 2, 3.

We will assume that the probability of getting heads and tails is the same:


p = q =1/2\def\arraystretch{1.5} \begin{array}{c:c} Possible \ Outcomes & A \\ \hline DDD & 3 \\ \hdashline DDN & 2 \\ \hdashline DND & 2 \\ \hdashline NDD & 2 \\ \hdashline DNN& 1 \\ \hdashline NDN & 1 \\ \hdashline NND & 1 \\ \hdashline NNN & 0 \\ \hdashline \end{array}



c. Construct the probability distribution of the random variable


\def\arraystretch{1.5} \begin{array}{c:c} x & 0 & 1 & 2 & 3 \\ \hline p(x) & 1/8 & 3/8 & 3/8 & 1/8 \end{array}

or


\def\arraystretch{1.5} \begin{array}{c:c} x & 0 & 1 & 2 & 3 \\ \hline p(x) & 0.125 & 0.375 & 0.375 & 0.125 \end{array}



d.

P(X=0)+P(X=1)+P(X=2)+P(X=3)

=1/8+3/8+3/8+1/8=1

e.

The probability of each value of a discrete random variable is between 0 and 1, and the sum of all the probabilities is equal to 1.

a.There are 2^3=8 possible outcomes


S=\{DDD, DDN, DND, NDD,

DNN,NDN, NND, NNN\}


b. The possible values of the random variable X are 0, 1, 2, 3.

We will assume that the probability of getting heads and tails is the same:


p = q =1/2\def\arraystretch{1.5} \begin{array}{c:c} Possible \ Outcomes & A \\ \hline DDD & 3 \\ \hdashline DDN & 2 \\ \hdashline DND & 2 \\ \hdashline NDD & 2 \\ \hdashline DNN& 1 \\ \hdashline NDN & 1 \\ \hdashline NND & 1 \\ \hdashline NNN & 0 \\ \hdashline \end{array}



c. Construct the probability distribution of the random variable


\def\arraystretch{1.5} \begin{array}{c:c} x & 0 & 1 & 2 & 3 \\ \hline p(x) & 1/8 & 3/8 & 3/8 & 1/8 \end{array}

or


\def\arraystretch{1.5} \begin{array}{c:c} x & 0 & 1 & 2 & 3 \\ \hline p(x) & 0.125 & 0.375 & 0.375 & 0.125 \end{array}



d.

P(X=0)+P(X=1)+P(X=2)+P(X=3)

=1/8+3/8+3/8+1/8=1

e.

The probability of each value of a discrete random variable is between 0 and 1, and the sum of all the probabilities is equal to 1.

a.There are 2^3=8 possible outcomes


S=\{DDD, DDN, DND, NDD,

DNN,NDN, NND, NNN\}


b. The possible values of the random variable X are 0, 1, 2, 3.

We will assume that the probability of getting heads and tails is the same:


p = q =1/2\def\arraystretch{1.5} \begin{array}{c:c} Possible \ Outcomes & A \\ \hline DDD & 3 \\ \hdashline DDN & 2 \\ \hdashline DND & 2 \\ \hdashline NDD & 2 \\ \hdashline DNN& 1 \\ \hdashline NDN & 1 \\ \hdashline NND & 1 \\ \hdashline NNN & 0 \\ \hdashline \end{array}



c. Construct the probability distribution of the random variable


\def\arraystretch{1.5} \begin{array}{c:c} x & 0 & 1 & 2 & 3 \\ \hline p(x) & 1/8 & 3/8 & 3/8 & 1/8 \end{array}

or


\def\arraystretch{1.5} \begin{array}{c:c} x & 0 & 1 & 2 & 3 \\ \hline p(x) & 0.125 & 0.375 & 0.375 & 0.125 \end{array}



d.

P(X=0)+P(X=1)+P(X=2)+P(X=3)

=1/8+3/8+3/8+1/8=1

e.

The probability of each value of a discrete random variable is between 0 and 1, and the sum of all the probabilities is equal to 1.


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