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Sedaia [141]
3 years ago
10

An urn contains 3 red and 7 black balls. players a and b withdraw balls from the urn consecutively until a red ball is selected.

find the probability that a selects the red ball. (a draws the first ball, then b, and so on. there is no replacement of the balls drawn.)
Mathematics
1 answer:
Elza [17]3 years ago
7 0

An urn contains 3 red and 7 black balls (10 in total). The probability to withdraw by player a first ball red from the urn is 3/10. If he draws black ball (the probability then is 1-3/10=7/10) , then player b must withdraw second ball black. In the urn remain 9 balls (among them 6 black). The probability to choose black ball is 7/10·6/9. Then if player a select a red ball, the probability becomes 7/10·6/9·3/8 (only 8 balls left and 3 red among them) and the probability for the player a to seelct black ball is 7/10·6/9·5/8 (only 8 balls left and 5 black among them) and so on;

Player a: 3/10, 7/10·6/9·3/8, 7/10·6/9·5/8·4/7·3/6, 7/10·6/9·5/8·4/7·3/6·2/5·3/4.

Use the sum rule to calculate the total probability;

3/10+ 7/10·6/9·3/8+ 7/10·6/9·5/8·4/7·3/6+ 7/10·6/9·5/8·4/7·3/6·2/5·3/4

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How many numbers are 10 units from zero on number line
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5 0
3 years ago
68 miles to 42.5 is what percent of decrease??
KatRina [158]
68 miles to 42.5's percent of decrease is: 37.5%
Using proportions. 68 miles - 42.5 = 25.5

25.5 x
------ = -----
68 100

Cross multiply:
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6 0
3 years ago
The mean cost of a five pound bag of shrimp is 50 dollars with a variance of 64. If a sample of 43 bags of shrimp is randomly se
anyanavicka [17]

Answer:

0.4122 = 41.22% probability that the sample mean would differ from the true mean by greater than 1 dollar

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation(which is the square root of the variance) \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, the sample means with size n of at least 30 can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 50, \sigma = \sqrt{64} = 8, n = 43, s = \frac{8}{\sqrt{43}} = 1.22

What is the probability that the sample mean would differ from the true mean by greater than 1 dollar?

Either it differs by 1 dollar or less, or it differs by more than one dollar. The sum of the probabilities of these events is decimal 1.

Probability it differs by 1 dollar or less:

pvalue of Z when X = 50+1 = 51 subtracted by the pvalue of Z when X = 50 - 1 = 49.

X = 51

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{51-50}{1.22}

Z = 0.82

Z = 0.82 has a pvalue of 0.7939

X = 49

Z = \frac{X - \mu}{s}

Z = \frac{49-50}{1.22}

Z = -0.82

Z = -0.82 has a pvalue of 0.2061

0.7939 - 0.2061 = 0.5878

Probability it differs by more than 1 dollar:

p + 0.5878 = 1

p = 0.4122

0.4122 = 41.22% probability that the sample mean would differ from the true mean by greater than 1 dollar

4 0
3 years ago
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