Mathematics, 24.03.2020 19:44 brinson1k

6. (4.2.12) Of the items manufactured by a certain process, 20% are defective. Of the defective items, 60% can be repaired. a. Find the probability that a randomly chosen item is defective and cannot be repaired. b. Find the probability that exactly 2 of 20 randomly chosen items are defective and cannot be repaired.

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Answer from: laraejackson75

(a) Probability that a randomly chosen item is defective and cannot be repaired is 8%.

(b) Probability that exactly 2 of 20 randomly chosen items are defective and cannot be repaired is 0.2711.

Step-by-step explanation:

We are given that of the items manufactured by a certain process, 20% are defective. Of the defective items, 60% can be repaired.

Let Probability that item are defective = P(D) = 0.20

Also, R = event of item being repaired

Probability of items being repaired from the given defective items = P(R/D) = 0.60

So, Probability of items not being repaired from the given defective items = P(R'/D) = 1 - P(R/D) = 1 - 0.60 = 0.40

(a) Probability that a randomly chosen item is defective and cannot be repaired = Probability of items being defective $\times$ Probability of items not being repaired from the given defective items

= 0.20 $\times$ 0.40 = 0.08 or 8%

So, probability that a randomly chosen item is defective and cannot be repaired is 8%.

(b) Now we have to find the probability that exactly 2 of 20 randomly chosen items are defective and cannot be repaired.

The above situation can be represented through Binomial distribution;

$P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....$

where, n = number of trials (samples) taken = 20 items

r = number of success = exactly 2

p = probability of success which in our question is % of randomly

chosen item to be defective and cannot be repaired, i.e; 8%

LET X = Number of items that are defective and cannot be repaired

So, it means X ~ Binom(n=20, p=0.08)

Now, Probability that exactly 2 of 20 randomly chosen items are defective and cannot be repaired is given by = P(X = 2)

P(X = 2) = $\binom{20}{2} \times 0.08^{2} \times (1-0.08)^{20-2}$

= $190 \times 0.08^{2} \times 0.92^{18}$

= 0.2711

Therefore, probability that exactly 2 of 20 randomly chosen items are defective and cannot be repaired is 0.2711.

Answer from: Quest

wholesale price is the initial price before markup. retail price comes after the markup.

a price of 8 bucks that's marked up by 100% is

formula: initial price + (percent markup in fraction/decimal)(initial price)

8+(1)(8)

8+8

retail price is $16.

Answer from: Quest

step-by-step explanation:

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