Question

Questions A and B included in the photo C) determine the shape of the probability distribution of x (left skewed, right skewed, symmetric, uniform) D) find the mean number of aces drawn when three cards are randomly selected E) find the standard deviation for the number of aces drawn when three cards are randomly selected F) find the probability of drawing one or less aces

Answer 1

(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

`Pr(\text{ace)}=(4)/(52)=(1)/(13)`

The probability of success and failure in this case is shown below

`\begin{gathered} p\Rightarrow\text{probability of success} \\ q\Rightarrow\text{probability of failure} \\ p\Rightarrow Pr(\text{ace)}=(1)/(13) \\ q\Rightarrow1-p=1-(1)/(13)=(12)/(13) \end{gathered}`

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

`Pr(r=x)=^nC_r* p^r* 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}`

For x=0

`\begin{gathered} r=x=0,p=(1)/(13),q=(12)/(13),n=3 \\ Pr(r=0)=^3C_0*((1)/(13))^0*((12)/(13))^3 \\ =1*1*0.7865=0.7865 \end{gathered}`

For x=1

`\begin{gathered} r=x=1,p=(1)/(13),q=(12)/(13),n=3 \\ Pr(r=1)=^3C_1*((1)/(13))^1*((12)/(13))^2 \\ =3*(1)/(13)*(144)/(169) \\ =(432)/(2197)=0.1966 \end{gathered}`

For x=2

`\begin{gathered} r=x=2,p=(1)/(13),q=(12)/(13),n=3 \\ Pr(r=2)=^3C_2*((1)/(13))^2*((12)/(13))^1 \\ =3*(1)/(169)*(12)/(13) \\ =(36)/(2197)=0.0164 \end{gathered}`

For x=3

`\begin{gathered} r=x=3,p=(1)/(13),q=(12)/(13),n=3 \\ Pr(r=2)=^3C_3*((1)/(13))^3*((12)/(13))^0 \\ =1*(1)/(2197)*1 \\ =(1)/(2197)=0.0005 \end{gathered}`

Answer: The probability distribution table is shown below