1. ## Probability question

I'm studying for a test and having trouble with this review question. Would you use binomial distribution on this? So convert your 2% to a faction 1/50 and then finish the binomial distribution. I see how that would work for a) but not sure about b) and c). Any guidance would be appreciated thanks.

A light bulb manufacture claims that 2% of their light bulbs are thrown out due to defects. If you sample 50 light bulbs, what is the probability of finding :
a) 3 defective?
b) 2 or less defective?
c) More than 3 defective?

2. Originally Posted by action259
I'm studying for a test and having trouble with this review question. Would you use binomial distribution on this? So convert your 2% to a faction 1/50 and then finish the binomial distribution. I see how that would work for a) but not sure about b) and c). Any guidance would be appreciated thanks.

A light bulb manufacture claims that 2% of their light bulbs are thrown out due to defects. If you sample 50 light bulbs, what is the probability of finding :
a) 3 defective?
b) 2 or less defective?
c) More than 3 defective?
I am just learning the binomial distribution in class now, so I will have a go at your questions.

I would use a binomial for this question, definitely. As you have a fixed number of trials, and a fixed probability of success/failure

Let X = the number of defective light bulbs in a sample of 50,

than X~B(50, 0.02)

than just use the formula, as you cant use the tables. $P(X=x) =\displaystyle \binom{n}{x} p^{x} (1-p)^{n-x}$

P(X=3) ?
$P(X \leq2)$ ?
P(X>3)?

3. Hello, action259!

A light bulb manufacture claims that 2% of their light bulbs are defective.
We have: . $\begin{Bmatrix}P(\text{d{e}f.}) &=& \frac{1}{50} \\ \\[-3mm] P(\text{good}) &=& \frac{49}{50} \end{Bmatrix}$

If you sample 50 light bulbs, what is the probability of finding:

a) 3 defective?
$P(\text{3 d{e}f}) \;=\;{50\choose3}\left(\frac{1}{50}\right)^3\left( \frac{49}{50}\right)^{47}$

b) 2 or less defective?
"2 or less" means: .0 def or 1 def or 2 def.

. . $P(\text{0 d{e}f}) \;=\;{50\choose0}\left(\frac{1}{50}\right)^0\left( \frac{49}{50}\right)^{50}$

. . $P(\text{1 d{e}f}) \;=\;{50\choose1}\left(\frac{1}{50}\right)\left(\f rac{49}{50}\right)^{49}$

. . $P(\text{2 d{e}f}) \;=\;{50\choose2}\left(\frac{1}{50}\right)^2\left( \frac{49}{50}\right)^{48}$

The answer is the sum of the three probabilities.

. . $P(\text{2 or less}) \;=\;P(\text{0 d{e}f}) + P(\text{1 d{e}f}) + P(\text{2 d{e}f})$

c) More than 3 defective?
The opposite of "more than 3" is "3 or less."

$P(\text{3 or less}) \:=\:\underbrace{P(\text{0 d{e}f}) + P(\text{1 d{e}f}) + P(\text{2 d{e}f})}_{\text{We found this in part (b)}} \:+\: P(\text{3 d{e}f})$

So we need: . $P(\text{3 d{e}f}) \:=\:{50\choose3}\left(\frac{1}{50}\right)^3\left( \frac{49}{50}\right)^{47}$