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3 years ago

If you roll six sided dice what are the odds of rolling doubles

Mathematics

1 answer:

Ludmilka [50]3 years ago

5 0

Answer: 16.7%

There are 6 ways we can roll doubles out of a possible 36 rolls (6 x 6), for a probability of 6/36, or 1/6, on any roll of two fair dice. So you have a 16.7% probability of rolling doubles with 2 fair six-sided dice.

Second Answer : 6/36

Out of the 6^2 = 36 outcomes, there is only 6 that are doubles (double 1,double 2,… double 6). Therefore the probability is 6/36 i 1/6.

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4.37) A set of data whose histogram is extremely skewed yields a mean and standard deviation of 70 and 12, respectively. What is

Brums [2.3K]

Answer:

75%

88.89%

Step-by-step explanation:

Given :

Mean = 70

Standard deviation = 12

Since the data is said to be extremely skewed, we apply Chebyshev's theorem rather than the empirical rule :

The minimum proportion of observation between 46 and 94

Chebyshev's theorem :

1 - 1 / k²

k = number of standard deviations from the mean

k = (94 - 70) / 12 = 24 / 12 = 2

Hence, we have ;

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1 - 0.25 = 0.75

Hence, The minimum proportion of observation between 46 and 94 is 75%

Between 36 and 106 :

k = (106 - 70) / 12 ;

k = 36/12 = 3

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1 - 1/3² = 1 - 1/9 = 8/9 = 0.8888 = 88.89%

The minimum proportion of observation between 34 and 106 is 88.89%

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2 years ago

A company that produces fine crystal knows from experience that 13% of its goblets have cosmetic flaws and must be classified as

Kisachek [45]

Answer:

(a) The probability that only one goblet is a second among six randomly selected goblets is 0.3888.

(b) The probability that at least two goblet is a second among six randomly selected goblets is 0.1776.

Step-by-step explanation:

Let X = number of seconds in the batch.

The probability of the random variable X is, p = 0.31.

The random variable X follows a Binomial distribution with parameters n and p.

The probability mass function of X is:

$P(X=x)={n\choose x}p^{x}(1-p)^{n-x};\ x=0,1,2,3...$

(a)

Compute the probability that only one goblet is a second among six randomly selected goblets as follows:

$P(X=1)={6\choose 1}0.13^{1}(1-0.13)^{6-1}\\=6\times 0.13\times 0.4984\\=0.3888$

Thus, the probability that only one goblet is a second among six randomly selected goblets is 0.3888.

(b)

Compute the probability that at least two goblet is a second among six randomly selected goblets as follows:

P (X ≥ 2) = 1 - P (X < 2)

$=1-{6\choose 0}0.13^{0}(1-0.13)^{6-0}-{6\choose 1}0.13^{1}(1-0.13)^{6-1}\\=1-0.4336+0.3888\\=0.1776$

Thus, the probability that at least two goblet is a second among six randomly selected goblets is 0.1776.

(c)

If goblets are examined one by one then to find four that are not seconds we need to select either 4 goblets that are not seconds or 5 goblets including only 1 second.

P (4 not seconds) = P (X = 0; n = 4) + P (X = 1; n = 5)

$={4\choose 0}0.13^{0}(1-0.13)^{4-0}+{5\choose 1}0.13^{1}(1-0.13)^{5-1}\\=0.5729+0.3724\\=0.9453$

Thus, the probability that at most five must be selected to find four that are not seconds is 0.9453.

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3 years ago