Exactly two dead batteries in three

Question

A drawer holds 9 batteries, of which 3 are dead and 6 still work. You pull out 3 batteries at random. What is the probability that exactly 2 of the 3 are dead?

Accepted Answer

Pulling 3 at once is an unordered pick, so count groups with combinations, where \binom{n}{k} is the number of ways to choose k from n. Total ways to choose 3 batteries from 9: \binom{9}{3}=84. Favorable: exactly 2 dead and 1 working. Pick 2 of the 3 dead and 1 of the 6 working: \binom{3}{2}	imes\binom{6}{1}=3	imes 6=18. Probability: \frac{18}{84}=\frac{3}{14}\approx 0.214. Takeaway. With only 3 dead batteries available, landing exactly 2 of them is possible but not common. The combination split makes the count routine.

Exactly two dead batteries in three

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