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Dennis_Churaev [7]
4 years ago
6

Find the probability that 3 randomly selected people all have the same birthday. ignore leap years. round to eight decimal place

s.\
Mathematics
2 answers:
IgorLugansk [536]4 years ago
7 0

Answer:

Probability that 3 randomly selected persons have same birthday = 0.00820417

Step-by-step explanation:

Probability that 1 person has different birthday = 1

\text{Probability that 2nd person has different birthday than 1st = }\frac{364}{365}\\\\\text{Probability that the 3rd person has different from 1st and 2nd = }\frac{363}{365}

\text{Probability that the all three persons have different birthday = }1\times \frac{364}{365}\times \frac{363}{365}\\\\=0.99179583

The probability that all three persons have same birthday = 1 - 0.99179583

                                                                                                 = 0.00820417

Hence, probability that 3 randomly selected persons have same birthday = 0.00820417

Alisiya [41]4 years ago
6 0

The probability the 1st person has a birthday is 365/365 = 1
The probability the 2nd person has a DIFFERENT birthday is 364/365 = 0.9972603
The probability the 3rd person also has a DIFFERENT birthday is 364/365 = 0.9972603
 

the probability that they all have the same birthday is

1 - (0.9972603)(0.9972603) = 0.00547189

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For purely rational functions, the general strategy is to compare the degrees of the numerator and denominator.

A)

\displaystyle \lim_{x\to\infty} \frac{2x^2-5}{7x^2+x-3} = \boxed{\frac27}

because both numerator and denominator have the same degree (2), so their end behaviors are similar enough that the ratio of their coefficients determine the limit at infinity.

More precisely, we can divide through the expression uniformly by <em>x</em> ²,

\displaystyle \lim_{x\to\infty} \frac{2x^2-5}{7x^2+x-3} = \lim_{x\to\infty} \frac{2-\dfrac5{x^2}}{7+\dfrac1x-\dfrac3{x^2}}

Then each remaining rational term converges to 0 as <em>x</em> gets arbitrarily large, leaving 2 in the numerator and 7 in the denominator.

B) By the same reasoning,

\displaystyle \lim_{x\to\infty} \frac{5x-3}{2x+1} = \boxed{\frac52}

C) This time, the degree of the denominator exceeds the degree of the numerator, so it grows faster than <em>x</em> - 1. Dividing a number by a larger number makes for a smaller number. This means the limit will be 0:

\displaystyle \lim_{x\to-\infty} \frac{x-1}{x^2+8} = \boxed{0}

More precisely,

\displaystyle \lim_{x\to-\infty} \frac{x-1}{x^2+8} = \lim_{x\to-\infty}\frac{\dfrac1x-\dfrac1{x^2}}{1+\dfrac8{x^2}} = \dfrac01 = 0

D) Looks like this limit should read

\displaystyle \lim_{t\to\infty}\frac{\sqrt{t}+t^2}{3t-t^2}

which is just another case of (A) and (B); the limit would be

\displaystyle \lim_{t\to\infty}\frac{\sqrt{t}+t^2}{3t-t^2} = -1

That is,

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However, in case you meant something else, such as

\displaystyle \lim_{t\to\infty}\frac{\sqrt{t+t^2}}{3t-t^2}

then the limit would be different:

\displaystyle \lim_{t\to\infty}\frac{\sqrt{t^2}\sqrt{\dfrac1t+1}}{3t-t^2} = \lim_{t\to\infty}\frac{t\sqrt{\dfrac1t+1}}{3t-t^2} = \lim_{t\to\infty}\frac{\sqrt{\dfrac1t+1}}{3-t} = 0

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One important detail glossed over here is that

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