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3 years ago

Find the probability that you roll the same number twice in a row when you toss two fair 6-sided dice.

Mathematics

1 answer:

marysya [2.9K]3 years ago

8 0

Answer: 6/36

Step-by-step explanation:

When tossing two number cubes, you can get such sample space

(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)

(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)

(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)

(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)

(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)

(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)

There are 36 possible outcomes. Only (1,1), (2,2), (3,3), (4,4), (5,5), and (6,6) use the same number twice.

Hence, the probability that you roll two numbers that are the same in a row is 6/36, or in simplified form, 1/6.

Hope that helps! :)

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Read 2 more answers

You have 1 case of soap bars and there are 150 bars in a case. You use 300 bars of soap per day. How many cases do you need to o

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Answer:
14 cases

Explanation:
First, we will need to find the number of cases needed per day:
We know that one case includes 150 bars and that the daily consumption is 300 bars. To know the number of cases needed daily, we will use cross multiplication as follows:
1 case .............> 150 bars
?? cases .........> 300 bars
number of cases needed daily = $\frac{300*1}{150}$ 
number of cases needed daily = 2 cases

Then, we will get the number of cases needed weekly (7 days):
We know that 2 cases are needed daily. To know the number of cases needed weekly (7 days), we will simply use cross multiplication as follows:
1 day .............> 2 cases
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Hope this helps :)

5 0

3 years ago

Suppose that the times required for a cable company to fix cable problems in its customers’ homes are uniformly distributed betw

vampirchik [111]

Answer:

a) 60% probability that a randomly selected cable repair visit will take at least 50 minutes

b) 60% probability that a randomly selected cable repair visit will take at most 55 minutes.

c) The expected length of the repair visit is 52.5 minutes.

d) The standard deviation is 7.22 minutes.

Step-by-step explanation:

An uniform probability is a case of probability in which each outcome is equally as likely.

For this situation, we have a lower limit of the distribution that we call a and an upper limit that we call b.

The probability that we find a value X lower than x is given by the following formula.

$P(X \leq x) = \frac{x - a}{b-a}$

For this problem, we have that:

Uniformly distributed between 40 and 65 minutes, so $a = 40, b = 65$

(a) What is the probability that a randomly selected cable repair visit will take at least 50 minutes?

Either it takes less than 50 minutes, or it takes at least 50 minutes. The sum of the probabilities of these events is decimal 1. So

$P(X < 50) + P(X \geq 50) = 1$

The intervals can be open or closed, so

$P(X < 50) = P(X \leq 50)$

We have that

$P(X \geq 50) = 1 - P(X < 50)$

$P(X < x) = \frac{50 - 40}{65 - 40} = 0.4$

$P(X \geq 50) = 1 - P(X < 50) = 1 - 0.4 = 0.6$

60% probability that a randomly selected cable repair visit will take at least 50 minutes

(b) What is the probability that a randomly selected cable repair visit will take at most 55 minutes?

This is $P(X \leq 55)$

$P(X \leq 55) = \frac{55 - 40}{65 - 40} = 0.6$

60% probability that a randomly selected cable repair visit will take at most 55 minutes.

(c) What is the expected length of the repair visit?

The mean of the uniform distribution is:

$M = \frac{a+b}{2}$

$M = \frac{40+65}{2} = 52.5$

The expected length of the repair visit is 52.5 minutes.

(d) What is the standard deviation?

The standard deviation of the uniform distribution is

$S = \sqrt{\frac{(b-a)^{2}}{12}}$

$S = \sqrt{\frac{(65-40)^{2}}{12}}$

$S = 7.22$

The standard deviation is 7.22 minutes.

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