(a) The random variable in this experiment is the number of aces in the experiment, that is all possible values of x {0,1,2,3}
(b) Given that the number of standard deck of cards is 52 and there are 4 aces, the probability of picking an ace is
[tex]Pr(\text{ace)}=\frac{4}{52}=\frac{1}{13}[/tex]The probability of success and failure in this case is shown below
[tex]\begin{gathered} p\Rightarrow\text{probability of success} \\ q\Rightarrow\text{probability of failure} \\ p\Rightarrow Pr(\text{ace)}=\frac{1}{13} \\ q\Rightarrow1-p=1-\frac{1}{13}=\frac{12}{13} \end{gathered}[/tex]To construct a probability distribution table for the number of aces for x values equal to 0,1,2,3. We are going to use binomial distribution formula, the binomial distribution formula is shown below
[tex]Pr(r=x)=^nC_r\times p^r\times q^{n-r},\Rightarrow\begin{cases}n=\text{the }number\text{ of trials} \\ r=the\text{ number of }specific\text{ outcomes in the trial} \\ x=0,1,2,3\end{cases}[/tex]For x=0
[tex]\begin{gathered} r=x=0,p=\frac{1}{13},q=\frac{12}{13},n=3 \\ Pr(r=0)=^3C_0\times(\frac{1}{13})^0\times(\frac{12}{13})^3 \\ =1\times1\times0.7865=0.7865 \end{gathered}[/tex]For x=1
[tex]\begin{gathered} r=x=1,p=\frac{1}{13},q=\frac{12}{13},n=3 \\ Pr(r=1)=^3C_1\times(\frac{1}{13})^1\times(\frac{12}{13})^2 \\ =3\times\frac{1}{13}\times\frac{144}{169} \\ =\frac{432}{2197}=0.1966 \end{gathered}[/tex]For x=2
[tex]\begin{gathered} r=x=2,p=\frac{1}{13},q=\frac{12}{13},n=3 \\ Pr(r=2)=^3C_2\times(\frac{1}{13})^2\times(\frac{12}{13})^1 \\ =3\times\frac{1}{169}\times\frac{12}{13} \\ =\frac{36}{2197}=0.0164 \end{gathered}[/tex]For x=3
[tex]\begin{gathered} r=x=3,p=\frac{1}{13},q=\frac{12}{13},n=3 \\ Pr(r=2)=^3C_3\times(\frac{1}{13})^3\times(\frac{12}{13})^0 \\ =1\times\frac{1}{2197}\times1 \\ =\frac{1}{2197}=0.0005 \end{gathered}[/tex]Answer: The probability distribution table is shown below
