SOLUTION: Evertight, a leading manufacturer of quality nails, produces 1-, 2-, 3-, 4-, and 5-inch nails for various uses. In the production process, if there is an overrun or the nails are s

Question 1192923: Evertight, a leading manufacturer of quality nails, produces 1-, 2-, 3-, 4-, and 5-inch nails for various uses. In the production process, if there is an overrun or the nails are slightly defective, they are placed in a common bin. Yesterday, 651 of the 1-inch nails, 243 of the 2-inch nails, 41 of the 3-inch nails, 451 of the 4-inch nails, and 333 of the 5-inch nails were placed in the bin. (a) What is the probability of reaching into the bin and getting a 4-inch nail? (b) What is the probability of getting a 5-inch nail? (c) If a particular application requires a nail that is 3 inches or shorter, what is the probability of getting a nail that will sa
Found 2 solutions by ikleyn, math_tutor2020:
Answer by ikleyn(52787) (Show Source): You can put this solution on YOUR website!
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Evertight, a leading manufacturer of quality nails, produces 1-, 2-, 3-, 4-, and 5-inch nails
for various uses. In the production process, if there is an overrun or the nails are slightly defective,
they are placed in a common bin. Yesterday, 651 of the 1-inch nails, 243 of the 2-inch nails,
41 of the 3-inch nails, 451 of the 4-inch nails, and 333 of the 5-inch nails were placed in the bin.
(a) What is the probability of reaching into the bin and getting a 4-inch nail?
(b) What is the probability of getting a 5-inch nail?
(c) If a particular application requires a nail that is 3 inches or shorter,
what is the probability of getting a nail that will sa
~~~~~~~~~~~~~~~~~

(a) P = . (b) P = . (c) P = .

Everything is so clear in this problem that more detailed explanations are excessive.

The formulas are SELF-EXPLANATORY.

If you have questions, look and hear attentively what these formulas tell you.

Answer by math_tutor2020(3817) (Show Source): You can put this solution on YOUR website!

Part (a)

A = 651 = number of 1-inch nails
B = 243 = number of 2-inch nails
C = 41 = number of 3-inch nails
D = 451 = number of 4-inch nails
E = 333 = number of 5-inch nails

A+B+C+D+E = 651+243+41+451+333 = 1719 nails total

Of that total, D = 451 are four-inch nails.

Answer: 451/1719

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Part (b)

We use the same idea as part (a)

E = 333 = number of five inch nails
A+B+C+D+E = 1719 nails total

It leads to the fraction 333/1719 which reduces to 37/191 after dividing both parts by the GCF 9.

Answer: 37/191

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Part (c)

The question is cut off. Please repost. Thank you.

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