A small showroom has $50$ LED Monitors on a shelf that work perfectly and another $5$ that are defective in the same shelf. What is the probability of randomly selecting two defective monitors when purchasing a pair of LED monitors from that showroom?
$W=$ working monitors, $NW=$ defective monitors
$P(W)=\frac{50}{55}$, $P(NW)=\frac{5}{55}$
$P(2$ defective monitors$)= \frac{5}{55} \cdot \frac{4}{54} = \frac{2}{297}$
This is what I did. Is this correct or am I wrong somewhere?
On
I have not checked both your and my answers. But this is how I go about solving it: It's a counting combination problem. The probability of choosing $2$ defective monitors $ = \dfrac{\text{the number of ways to choose 2 defective ones}}{\text{the number of ways to choose two items}} = \dfrac{\binom{5}{2}}{\binom{55}{2}}=...$. I hope you can work it out the math.
Yes, your solution is correct.
The probability of picking a wrong monitor the first time is obviously $\frac{5}{55}$, and the probability for the second time is $\frac{5-1}{55-1}=\frac{4}{54}$, so you multiply those and get the result.
Ps: you can also solve it using $\dfrac{\binom{5}{2}}{\binom{55}{2}}$ as the other answer had shown. Anyway, the result is the same.