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julia-pushkina [17]
3 years ago
11

Five balls, numbered 1, 2, 3, 4, and 5, are placed in an urn. Two balls are randomly selected from the five, and their numbers n

oted. Find the probability distribution for the following: a The largest of the two sampled numbers b The sum of the two sampled numbers
Mathematics
1 answer:
nirvana33 [79]3 years ago
8 0

Answer

given,

five balls numbered 1, 2, 3, 4, and 5.

two balls are drawn randomly

number of ways of selecting two balls = ⁵C₂ = 10 ways

For sample to be of largest number there is two option 4 or 5

so for first section of ball probability will be = \dfrac{2}{5}

now one ball is selected so for the selection of second ball the ball are 4

probability of second ball selection = \dfrac{1}{4}  

a) largest of the two sample

               = \dfrac{2}{5}\times \dfrac{1}{4}

               = \dfrac{1}{10}

               =0.1

b) sum of two sample

               =  \dfrac{2}{5}+\dfrac{1}{4}  

               = \dfrac{13}{20}

               = 0.65

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Step-by-step explanation:

Given

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Defective phones = 9

So the probability of defective phones will be calculated by dividing the number of defective phones by total number of phones checked.

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= 0.3 or 30%

So, from 1000 phones the defective phones will be:

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Simplify the expression. Rewrite using only positive exponents.
Dvinal [7]

The simplified form of the given expression is x^{18}a^{3}b^{3}.

The given expression is (\frac{x^{4}a^{2}b^{3} }{x^{-2}ab^{2}} )^{3}.

We need to simplify the given expression.

<h3>What are the basic laws of exponents?</h3>

The basic laws of exponents are as follows:

a^{m} \times b^{m} =(ab)^{m}

a^{m} \div b^{m} =(\frac{a}{b} )^{m}

a^{m} \times a^{n} =(a)^{m+n}

a^{m} \div a^{n} =(a)^{m-n}

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=x^{18}a^{3}b^{3}

Therefore, the simplified form of the given expression is x^{18}a^{3}b^{3}.

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