Given:
Number of cameras = 10
The number defectives are 6 that means 4 are non defectives.
Now, persons buys 2 cameras.
To find the probability of getting 2 defectives,
[tex]\begin{gathered} P=\frac{Favorable\text{ ways}}{\text{total ways}} \\ =\frac{^6C_2\times^4C_0}{^{10}C_2_{}} \\ As^nC_r=\frac{n!}{r!(n-r)!}_{} \\ P=\frac{15\times1}{45} \\ P=\frac{1}{3} \\ P=0.3333 \end{gathered}[/tex]Answer: Probability is 0.3333
