Question

A company that produces DVD drives has a 12% defective rate. Let X represent the number of defectives in a random sample of 53 of their drives. a. What is the probability the sample will contain exactly 3 defective drives? Give your answer to four decimal places. b. What is the probability the sample will contain more than 3 defective drives? Give your answer to four decimal places. c. What is the probability the sample will contain less than 3 defective drives? Give your answer to four decimal places. d. What is the expected number of defective drives in the sample? Give your answer to two decimal places.

Answer 1

Answer:

a. P(X=3)=0,0678

b. P(X>3) = 0,8935

c. P(X<3) = 0,0386

d. Approximately 6 DVD drives are expected to be defective.

Step-by-step explanation:

Hello!

To calculate de probabilities, lets first establish the study variable

X: Number of defective DVD drives in a sample of 53 DVD drives.

This is a discrete variable type and has a binomial distribution.

Binomial criteria:

1. The number of observation of the trial is fixed (In this case n = 53)

2. Each observation in the trial is independent, this means that none of the trials will have an effect on the probability of the next trial (In this case, Each DVD drive observed does not affect on the probability of the next one being defective)

3. The probability of success in the same from one trial to another (In this case our "success" will be finding a defective DVD drive; ρ=0,12)

So X≈ Bi (n;ρ)

Where n represents the sample (n=53) and ρ is the probability of success (ρ=0.12)

a. What is the probability the sample will contain 3 defective drives?

Symbolically

P(X=3)

Now we can apply the formula $P(X)=\frac{n!}{(n-X)!X!}*p^{x}*q^{n-x}$

Where

n= sample

x= number of success trials

p= probabilitry of success

q= probability of failure (calculate it as 1 - p)

If you replace the formula

$P(X=3)=\frac{53!}{(53-3)!3!}*0,12^{3}*0,88^{53-3}$

P(X=3)=0,0678

b. What is the probability the sample will contain more than 3 defective DVD drives.

Symbolically

P(X>3)

This one is a little tricky to calculate. Saying P(X>3) is the same as saying P(X≥4), so you can calculate all probabilities from X=4 to X=53 and add them or you can apply the second theorem of Kolmogorov’s Axioms (P(S)=1) and rewrite it as

P(X>3) = 1 - P(X≤3)

Where the expresion P(X≤3) indicates the cummulative probabilities until 3, in other words in this expresion are included the probabilities of X=0, X=1, X=2 and X=3

Let's calculate this probabilities using the formula

P(X=0) = $\frac{53!}{(53-0)!0!}*0,12{0}*0,88^{53-0}$

P(X=0) = $1,14178*10^{-3}$

P(X=1) = $\frac{53!}{(53-1)!1!}*0,12^{1}*0,88^{53-1}$

P(X=1) = $8,2519*10^{-3}$

P(X=2) = $\frac{53!}{(53-2)!2!}*0,12^{2}*0,88^{53-2}$

P(X=2) = 0,0292

And the P(X=3) was calculated in a.

P(X=3) = 0,0678

So P(X≤3) = P(X=0) + P(X=1) + P(X=2) + P(X=3)

P(X≤3) = $1,14177*10^{-3}$ + $8,2519*10^{-3}$ + 0,0292 + 0,0678

P(X≤3) = 0,1065

Now we can calculate the P(X>3)

P(X>3) = 1 - P(X≤3) = 1- 0,1065 = 0,8935

P(X>3) = 0,8935

c. What is the probability the sample will contain less than 3 defective drives?

Symbolically

P(X<3) = P(X≤2) = P(X=0) + P(X=1) + P(X=2)

Using the probabilities calculated in a. and b.

P(X<3) = $1,14177*10^{-3}$ + $8,2519*10^{-3}$ + 0,0292

P(X<3) = 0,0386

d. What is the expected number of defective drives in the sample?

This question is rather simple, to calculate the expected number of defective drivers (x) we use the probability of finding a defective drive (p) and the sample (n)

x = n * p = 53 * 0,12 = 6,36 DVD drives are expected to be defective approx 6 DVD drives.

I hope you have a SUPER day!