In a box of 7 batteries,3 are dead.You choose two batteries at random from the box.Let the random variable X be the number of good batteries you get.Find the probability model for X.
A) $$\begin{array} { l | c c c }
\text { Number good } & 0 & 1 & 2 \
\hline \mathrm { P } \text { (Number good) } & 0.184 & 0.490 & 0.327
\end{array}$$
B) $$\begin{array} { l | c c c }
\text { Number good } & 0 & 1 & 2 \
\hline \mathrm { P } \text { (Number good) } & 0.143 & 0.571 & 0.286
\end{array}$$
C) $$\begin{array} { l | c c c }
\text { Number good } & 0 & 1 & 2 \
\hline \mathrm { P } \text { (Number good) } & 0.143 & 0.286 & 0.286
\end{array}$$
D) $$\begin{array} { l | c c c }
\text { Number good } & 0 & 1 & 2 \
\hline \mathrm { P } \text { (Number good) } & 0.286 & 0.571 & 0.286
\end{array}$$
E) $$\begin{array} { l | c c c }
\text { Number good } & 0 & 1 & 2 \
\hline \mathrm { P } \text { (Number good) } & 0.067 & 0.467 & 0.467
\end{array}$$

Question

In a box of 7 batteries,3 are dead.You choose two batteries at random from the box.Let the random variable X be the number of good batteries you get.Find the probability model for X.
A) $$\begin{array} { l | c c c } 
\text { Number good } & 0 & 1 & 2 \
\hline \mathrm { P } \text { (Number good) } & 0.184 & 0.490 & 0.327
\end{array}$$
B) $$\begin{array} { l | c c c } 
\text { Number good } & 0 & 1 & 2 \
\hline \mathrm { P } \text { (Number good) } & 0.143 & 0.571 & 0.286
\end{array}$$
C) $$\begin{array} { l | c c c } 
\text { Number good } & 0 & 1 & 2 \
\hline \mathrm { P } \text { (Number good) } & 0.143 & 0.286 & 0.286
\end{array}$$
D) $$\begin{array} { l | c c c } 
\text { Number good } & 0 & 1 & 2 \
\hline \mathrm { P } \text { (Number good) } & 0.286 & 0.571 & 0.286
\end{array}$$
E) $$\begin{array} { l | c c c } 
\text { Number good } & 0 & 1 & 2 \
\hline \mathrm { P } \text { (Number good) } & 0.067 & 0.467 & 0.467
\end{array}$$

Quizplus · Accepted Answer

When choosing two batteries at random, there are a total of (7 choose 2) = 21 possible outcomes. 
We can find the probability of each possible value of X as follows:

- X = 0: There is only 1 way to choose 2 dead batteries out of 3, so P(X=0) = (3 choose 2)/(7 choose 2) = 3/21.
- X = 1: There are 2 ways to choose 1 good battery and 1 dead battery (either pick the two good ones or pick one good and one dead), so P(X=1) = 2*(4 choose 1)*(3 choose 1)/(7 choose 2) = 24/21.
- X = 2: There is only 1 way to choose 2 good batteries out of 4, so P(X=2) = (4 choose 2)/(7 choose 2) = 6/21.

Thus, the probability model for X is:
P(X=0) = 3/21
P(X=1) = 24/21
P(X=2) = 6/21

Note that the probabilities add up to 1, as they should for any probability model.

In a Box of 7 Batteries,3 Are Dead Number good 012P (Number good) 0.1840.4900.327\begin{array} { l | c c c } \text { Number good } & 0 & 1 & 2 \\\hline \mathrm { P } \text { (Number good) } & 0.184 & 0.490 & 0.327\end{array} Number good P (Number good) ​00.184​10.490​20.327​​

In a Box of 7 Batteries,3 Are Dead $\begin{array} { l | c c c } \text { Number good } & 0 & 1 & 2 \\\hline \mathrm { P } \text { (Number good) } & 0.184 & 0.490 & 0.327\end{array}$