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2 years ago

Consider two people being randomly selected. (For simplicity, ignore leap years.)

Mathematics

1 answer:

Setler79 [48]2 years ago

8 0

Answer:

A)144/133225

B)1/365

Step-by-step explanation:

A) We want to find the probability that two people have a birthday on the 18th of any month.

If we Neglect leap year like the question says, it means that there are 365 days in a year.

Now, we have 12 possible number of 18th day in all months that make a calendar year.

Thus;

If two people have birthday on 18th, the probability that the two of them will have birthday on the 18th of any month is;

P(both to have birthday on the 18th of any month) = (12/365) × (12/365) = 144/133225

B) We want to find the probability that two people have a birthday on the same day of the same month. This can be written as;

P(2 people have a birthday on the same day of the same month) + P(2 people do not having birthday on same day of same month) = 1

Probability of 2 people not having birthday on same day of same month = 365/365 × 364/365 = 364/365

P(2 people have a birthday on the same day of the same month) = 1 - (364/365) = 1/365

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3 years ago

According to the article "Characterizing the Severity and Risk of Drought in the Poudre River, Colorado" (J. of Water Res. Plann

mihalych1998 [28]

Answer:

(a) P (Y = 3) = 0.0844, P (Y ≤ 3) = 0.8780

(b) The probability that the length of a drought exceeds its mean value by at least one standard deviation is 0.2064.

Step-by-step explanation:

The random variable Y is defined as the number of consecutive time intervals in which the water supply remains below a critical value y₀.

The random variable Y follows a Geometric distribution with parameter p = 0.409.

The probability mass function of a Geometric distribution is:

$P(Y=y)=(1-p)^{y}p;\ y=0,12...$

(a)

Compute the probability that a drought lasts exactly 3 intervals as follows:

$P(Y=3)=(1-0.409)^{3}\times 0.409=0.0844279\approx0.0844$

Thus, the probability that a drought lasts exactly 3 intervals is 0.0844.

Compute the probability that a drought lasts at most 3 intervals as follows:

P (Y ≤ 3) = P (Y = 0) + P (Y = 1) + P (Y = 2) + P (Y = 3)

$=(1-0.409)^{0}\times 0.409+(1-0.409)^{1}\times 0.409+(1-0.409)^{2}\times 0.409\\+(1-0.409)^{3}\times 0.409\\=0.409+0.2417+0.1429+0.0844\\=0.8780$

Thus, the probability that a drought lasts at most 3 intervals is 0.8780.

(b)

Compute the mean of the random variable Y as follows:

$\mu=\frac{1-p}{p}=\frac{1-0.409}{0.409}=1.445$

Compute the standard deviation of the random variable Y as follows:

$\sigma=\sqrt{\frac{1-p}{p^{2}}}=\sqrt{\frac{1-0.409}{(0.409)^{2}}}=1.88$

The probability that the length of a drought exceeds its mean value by at least one standard deviation is:

P (Y ≥ μ + σ) = P (Y ≥ 1.445 + 1.88)

= P (Y ≥ 3.325)

= P (Y ≥ 3)

= 1 - P (Y < 3)

= 1 - P (X = 0) - P (X = 1) - P (X = 2)

$=1-[(1-0.409)^{0}\times 0.409+(1-0.409)^{1}\times 0.409\\+(1-0.409)^{2}\times 0.409]\\=1-[0.409+0.2417+0.1429]\\=0.2064$

Thus, the probability that the length of a drought exceeds its mean value by at least one standard deviation is 0.2064.

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3 years ago

Skip count 89300 forward by 5000

gtnhenbr [62]

89300, 94300, 99300, 104300, 109300. To keep going forward, add 5000 after each number.

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3 years ago