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s344n2d4d5 [400]
3 years ago
15

A box contains five slips of paper, each with one of the letters A, B, C, D, or E written on it. Another box has four slips of p

aper, each with one of the numbers 1, 2, 3, or 4 written on it. Without looking, Nancy draws a slip of paper from each box. What is the probability that Nancy draws the letter B from one box and the number 2 from the other box?
Mathematics
2 answers:
taurus [48]3 years ago
6 0

Answer:

The probability that Nancy draws the letter B from one box and the number 2 from the other box is:

\dfrac{1}{20}

Step-by-step explanation:

We know that the probability of drawing a slip of paper from one box is independent of drawing a piece of slip from the other.

Let first box contain five slips of paper, each with one of the letters A, B, C, D, or E written on it.

This means that the probability of drawing B is:

ratio of the number of slips containing B to the total number of slips.

i.e.

P(Drawing\ B)=\dfrac{1}{5}

Second box has four slips of paper, each with one of the numbers 1, 2, 3, or 4 written on it.

This means that the probability of drawing number 2 is:

ratio of the number of slips containing number 2 to the total number of slips.

i.e.

P(Drawing\ 2)=\dfrac{1}{4}

Hence,

P(drawing\ B\ and\ number\ 2)=P(drawing\ B)\times P(drawing\ 2)\\\\i.e.\\P(drawing\ B\ and\ number\ 2)=\dfrac{1}{5}\times \dfrac{1}{4}\\\\P(drawing\ B\ and\ number\ 2)=\dfrac{1}{20}

Hence, the probability is:

\dfrac{1}{20}

Alla [95]3 years ago
4 0
20% chance for the letter box and 25% chance for Number box.

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

1) Data given and notation

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X=18 represent the ounce cups of coffee that were underfilled

\hat p=\frac{18}{100}=0.18 estimated proportion of ounce cups of coffee that were underfilled

p_o=0.1 is the value that we want to test

\alpha represent the significance level  

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)  

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the true proportion it's higher than 0.1 or 10%:  

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Alternative hypothesis:p > 0.1  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

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The One-Sample Proportion Test is used to assess whether a population proportion \hat p is significantly different from a hypothesized value p_o.

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Since we have all the info requires we can replace in formula (1) like this:  

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The significance level assumed for this case is \alpha=0.05. The next step would be calculate the p value for this test.  

Since is a one right tailed test the p value would be:  

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So the p value obtained was a very low value and using the significance level given \alpha=0.05 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance thetrue proportion is not significanlty higher than 0.1 or 10% .  

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Step-by-step explanation:

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