Question

Grandkids Mr. Starnes and his wife have $6$grandchildren: Connor, Declan, Lucas, Piper, Sedona, and Zayne. They have $2$extra tickets to a holiday show, and will randomly select which $2$grandkids get to see the show with them.

a. Give a probability model for this chance process.

b. Find the probability that at least one of the two girls (Piper and Sedona) get to go to the show.

Short answer

Part(a) The probability model is $$\begin{gathered} \mathrm{X}~\mathrm{Bin}(6,\frac{1}{6}) \\ \mathrm{P}(\mathrm{X}=2)={}^6\mathrm{C}_2\times {\left(\frac{1}{6}\right)}^2\times {\left(1-\frac{1}{6}\right)}^{6-2} \end{gathered}$$

Part(b) The probability that at least one of the two girls (Piper and Sedona) get to go to the show is $0.6$

Step-by-step solution

Part(a) Step 1 : Given information

We need to give probability model for this chance process.

Part(a) Step 2 : Simplify

We are given that there are $6$grandchildrens and $2$of them are needed to picked.

Let number of grandchildren to be choose is X.

Every children has equal chance to get selected, therefore Probability for everyone is $\frac{1}{6}$

Therefore, the probability model is :

$$\begin{gathered} \mathrm{X}~\mathrm{Bin}(6,\frac{1}{6}) \\ \mathrm{P}\left(\mathrm{X}\right)={}^6\mathrm{C}_x\times {\left(\frac{1}{6}\right)}^x\times {\left(1-\frac{1}{6}\right)}^{6-x} \\ \mathrm{P}(\mathrm{X}=2)={}^6\mathrm{C}_2\times {\left(\frac{1}{6}\right)}^2\times {\left(1-\frac{1}{6}\right)}^{6-2} \end{gathered}$$

Part(b) Step 1 : Given information

We need to find probability for at least one of two girls get to go to the show.

Part(b) Step 2 : Simplify

As every children has equal chance to be selected , therefore

$$\begin{gathered} \mathrm{P}(\mathrm{at}\mathrm{least}\mathrm{one}\mathrm{of}\mathrm{the}\mathrm{two}\mathrm{girls})=\mathrm{P}(\mathrm{one}\mathrm{of}\mathrm{girls})+\mathrm{P}(\mathrm{both}\mathrm{girls}) \\ \mathrm{P}(\mathrm{one}\mathrm{of}\mathrm{girls})=\frac{{}^2\mathrm{C}_1\times {}^4\mathrm{C}_1}{{}^6\mathrm{C}_2}=0.5333 \\ \mathrm{P}(\mathrm{both}\mathrm{girls})=\frac{{}^2\mathrm{C}_2\times {}^4\mathrm{C}_0}{{}^6\mathrm{C}_2}=0.0667 \\ \mathrm{P}(\mathrm{at}\mathrm{least}\mathrm{one}\mathrm{of}\mathrm{the}\mathrm{two}\mathrm{girls})=0.5333+0.0667=0.6 \end{gathered}$$