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) $0.1840.4900.327\begin{array} { l | c c c } \text { Number good } & 0 & 1 & 2 \\\hline \mathrm { P } \text { (Number good) } & 0.184 & 0.490 & 0.327\end{array}$
B) $0.1430.5710.286\begin{array} { l | c c c } \text { Number good } & 0 & 1 & 2 \\\hline \mathrm { P } \text { (Number good) } & 0.143 & 0.571 & 0.286\end{array}$
C) $0.1430.2860.286\begin{array} { l | c c c } \text { Number good } & 0 & 1 & 2 \\\hline \mathrm { P } \text { (Number good) } & 0.143 & 0.286 & 0.286\end{array}$
D) $0.2860.5710.286\begin{array} { l | c c c } \text { Number good } & 0 & 1 & 2 \\\hline \mathrm { P } \text { (Number good) } & 0.286 & 0.571 & 0.286\end{array}$
E) $0.0670.4670.467\begin{array} { l | c c c } \text { Number good } & 0 & 1 & 2 \\\hline \mathrm { P } \text { (Number good) } & 0.067 & 0.467 & 0.467\end{array}$

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)     Number good  0 1 2 P  (Number good)   0.184 0.490 0.327 \begin{array} { l | c c c } 	ext { Number good }  &  0  &  1  &  2 \\hline \mathrm { P } 	ext { (Number good)  }  &  0.184  &  0.490  &  0.327\end{array}  Number good  P  (Number good)   ​ 0 0.184 ​ 1 0.490 ​ 2 0.327 ​ ​ B)     Number good  0 1 2 P  (Number good)   0.143 0.571 0.286 \begin{array} { l | c c c } 	ext { Number good }  &  0  &  1  &  2 \\hline \mathrm { P } 	ext { (Number good)  }  &  0.143  &  0.571  &  0.286\end{array}  Number good  P  (Number good)   ​ 0 0.143 ​ 1 0.571 ​ 2 0.286 ​ ​ C)     Number good  0 1 2 P  (Number good)   0.143 0.286 0.286 \begin{array} { l | c c c } 	ext { Number good }  &  0  &  1  &  2 \\hline \mathrm { P } 	ext { (Number good)  }  &  0.143  &  0.286  &  0.286\end{array}  Number good  P  (Number good)   ​ 0 0.143 ​ 1 0.286 ​ 2 0.286 ​ ​ D)     Number good  0 1 2 P  (Number good)   0.286 0.571 0.286 \begin{array} { l | c c c } 	ext { Number good }  &  0  &  1  &  2 \\hline \mathrm { P } 	ext { (Number good)  }  &  0.286  &  0.571  &  0.286\end{array}  Number good  P  (Number good)   ​ 0 0.286 ​ 1 0.571 ​ 2 0.286 ​ ​ E)     Number good  0 1 2 P  (Number good)   0.067 0.467 0.467 \begin{array} { l | c c c } 	ext { Number good }  &  0  &  1  &  2 \\hline \mathrm { P } 	ext { (Number good)  }  &  0.067  &  0.467  &  0.467\end{array}  Number good  P  (Number good)   ​ 0 0.067 ​ 1 0.467 ​ 2 0.467 ​ ​

adnan almaskari · Accepted Answer

Final Answer : B, Explanation :  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) = 3/21P(X=1) = 24/21P(X=2) = 6/21Note that the probabilities add up to 1, as they should for any probability model.

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