Answer: 0.167
Explanation:
Given: Total batteries = 10
Batteries that are still working = 6
Number of ways to pick 3 working batteries =
![^6C_3=(6!)/(3!3!)](https://img.qammunity.org/2021/formulas/mathematics/high-school/iixvthmzwspkk2sstl7ixw412zs406il0n.png)
![=(6*5*4*3!)/(6*3!)\\\\=20](https://img.qammunity.org/2021/formulas/mathematics/high-school/u6hj0kag0v8n32slom9d6g44y3fa2vi6zz.png)
Number of ways of pick 3 batteries out of 10 =
![^(10)C_3=(10!)/(3!7!)=(10*9*8*7!)/(6*7!)\\\\=120](https://img.qammunity.org/2021/formulas/mathematics/high-school/t5b7vrjj6a3mc7xcjm62qpiqmr0js6qzya.png)
Required probability =
![(20)/(120)=0.167](https://img.qammunity.org/2021/formulas/mathematics/high-school/wy7yhnq36ihkr4ql1hqtgqqd1z1est5g7r.png)
Hence, the probability that all of the first 3 she chooses will work = 0.167