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Dennis_Churaev [7]
3 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]3 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]3 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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5 0
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zlopas [31]

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<h2>B. 7.1</h2>

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S = \sqrt{\frac{(2-9.7)^2+(6-9.7)^2+(15-9.7)^2+(9-9.7)^2+(11-9.7)^2+(22-9.7)^2+(1-9.7)^2+(4-9.7)^2+(8-9.7)^2+(19-9.7)^2}{10-1} }\\ S = \sqrt{\frac{(-7.7)^2+(-3.7)^2+(5.3)^2+(-0.7)^2+(1.3)^2+(12.3)^2+(-8.7)^2+(-5.7)^2+(-1.7)^2+(9.3)^2}{10-1} }\\\\S =  \sqrt{\dfrac{59.29+13.69+28.09+0.49+1.69+151.29+75.69+32.49+2.89+86.49}{10-1} }\\\\\\S =  \sqrt{\dfrac{452.1}{9} }\\\\S = \sqrt{50.23}\\ \\S = 7.08\\\\S \approx 7.1

<em>Hence the standard deviation of the sample data is 7.1</em>

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3 years ago
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