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3 years ago

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In a box of 25 external hard disks, there are 2 defectives. An inspector examines 5 of these hard disks. Find the probability th

at there is at least 1 defective hard disk among the 5.

1 answer:

Salsk061 [2.6K]3 years ago

8 0

Answer:

0.3667 or 36.67%

Step-by-step explanation:

The probability of getting at least 1 defective hard disk among the 5 (P(X>0)) is equal to 100% minus the probability of getting none defectives (1-P(X=0)).

If 23 out of 25 hard disks are non-defective, the probability is:

$P(X>0) = 1 -P(X=0)\\P(X>0) = 1 -\frac{23}{25}*\frac{22}{24} *\frac{21}{23} *\frac{20}{22} *\frac{19}{21}\\ P(X>0) = 0.3667=36.67\%$

The probability that there is at least 1 defective hard disk is 0.3667 or 36.67%.

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Step-by-step explanation:

1.

Simplify the expression by combining like terms. Remember, like terms have the same variable part, to simplify these terms, one performs operations between the coefficients. Please note that a variable with an exponent is not the same as a variable without the exponent. A term with no variable part is referred to as a constant, constants are like terms.

$2h^2-7h+2h^2-h+6+4h-9h$

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2.

Use a very similar method to solve this problem as used in the first. Please note that all of the rules mentioned in the first problem also apply to this problem; for that matter, the rules mentioned in the first problem generally apply to any pre-algebra problem.

$8x^3y^2-7x^2y+8x-4-x^3y^2+2x^2y+4x^2y-8x+5$

$(8x^3y^2-x^3y^2)+(-7x^2y+2x^2y+4x^2y)+(8x-8x)+(-4+5)$

$7x^3y^2-x^2y+1$

3.

Use the same rules as applied in the first problem. Also, keep the distributive property in mind. In simple terms, the distributive property states the following ( $a(b+c)=(a)(b)+(a)(c)=ab+ac$ ). Also note, a term raised to an exponent is equal to the term times itself the number of times the exponent indicates. In the event that the term raised to an exponent is a constant, one can simplify it. Apply these properties here,

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$-2(8n+1)-(5-9n)+(3*3)$

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4.

The same method used to solve problem (3) can be applied to this problem.

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$\frac{1}{2}(10-8m+6m^2)-(3m^2+4m-7)-(2)(2)(2)$

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$(-3m^2-3m^2)+(-4m-4m)+(5+7-8)$

$-8m+4$

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2 years ago