Answer: 0.167
Step-by-step explanation:
Given: Total batteries = 10
Batteries that are still working = 6
Number of ways to pick 3 working batteries = [tex]^6C_3=\dfrac{6!}{3!3!}[/tex]
[tex]=\dfrac{6\times5\times4\times3!}{6\times3!}\\\\=20[/tex]
Number of ways of pick 3 batteries out of 10 =
[tex]^{10}C_3=\dfrac{10!}{3!7!}=\dfrac{10\times9\times8\times7!}{6\times7!}\\\\=120[/tex]
Required probability = [tex]\dfrac{20}{120}=0.167[/tex]
Hence, the probability that all of the first 3 she chooses will work = 0.167
