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Morgarella [4.7K]

3 years ago

8

A tray of electronic components contains 15 components, 4 of which are defective. If 4 components are selected, what is the poss

ibility that (a) all 4 are defective? (b) 3 are defective and 1 is good? (c) exactly 2 are defective? (d) none are defective?

1 answer:

dlinn [17]3 years ago

8 0

Answer:

a) 0.0007326

b) 0.03223

c) 0.2418

d) 0.2418

Explanation:

To find different probabilities for the selection of components among eleven good and four defective components, we will use the Combination.

a) $C(4,4) = 1; C(15,4) = 1365$

$P = \frac{C(4,4)}{C(15,4)} = \frac{1}{1365} = 0.0007326$

b) $C(4,3) = 4; C(11,1) = 11$

$P = \frac{C(4,3)*C(11,1)}{C(15,4)} = \frac{4*11}{1365} = 0.03223$

c) $C(4,2) = 6; C(11,2) = 55$

$P = \frac{C(4,2)*C(11,2)}{C(15,4)} = \frac{6*55}{1365} = 0.2418$

d) $C(11,4) = 330$

$P = \frac{C(11,4)}{C(15,4)} = \frac{330}{1365} = 0.2418$

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<u>Explanation:</u>

Given, Diameter of sphere = 6 cm

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Similarly,

$\text { wire radius, } r=\frac{0.2}{2}=0.1 \mathrm{mm}=1 \times 10^{-3} \mathrm{m}$

We know the below formulas,

$\text {volume of sphere}=\frac{4}{3} \times \pi \times R^{3}$

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When equating both the equations, we can find length of wire as below, where $\pi=\frac{22}{7}$

$\frac{4}{3} \times \pi \times R^{3}=\pi \times r^{2} \times l$

$\frac{4}{3} \times \frac{22}{7} \times\left(3 \times 10^{-2}\right)^{3}=\frac{22}{7} \times\left(1 \times 10^{-3}\right)^{2} \times l$

The $\pi$ value gets cancelled as common on both sides, we get

$\frac{4}{3} \times 27 \times 10^{-6}=10^{-6} \times l$

The $10^{-6}$ value gets cancelled as common on both sides, we get

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Answer:

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AlekseyPX

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B. The substances that are changed.

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ANSWER

The correct option is C.

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A man who works for a moving company is loading a box onto a moving van. He pushes a 200N box up a 5m long ramp. If he pushes wi

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