Question

A shipment of 10 cameras will likely have 6 defectives. If a person buys 2 cameras, what is the probability of getting 2 defectives?

Answer 1

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,

`\begin{gathered} P=\frac{Favorable\text{ ways}}{\text{total ways}} \\ =\frac{^6C_2*^4C_0}{^(10)C_2_{}} \\ As^nC_r=(n!)/(r!(n-r)!)_{} \\ P=(15*1)/(45) \\ P=(1)/(3) \\ P=0.3333 \end{gathered}`

Answer: Probability is 0.3333