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Aliun [14]
4 years ago
6

Find the probability that 4 randomly selected people all have the same birthday. Ignore leap years. Round to eight decimal place

s.
Mathematics
1 answer:
Ksju [112]4 years ago
7 0
The number of days of not leap years is 365.

So, the number of different variations of birthdays for 4 people is 365 * 365 * 365 * 365 = (365)^4

The probability that all of them be one specific day is 365 / (365)^4 = 1 / (365)^3 ≈ 0.00000002.

Answer: 0.00000002


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\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-10}~,~\stackrel{y_1}{1})\qquad \underline{(\stackrel{x_2}{-2}~,~\stackrel{y_2}{2})}\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ d=\sqrt{[-2-(-10)]^2+[2-1]^2}\implies d=\sqrt{(-2+10)^2+(2-1)^2} \\\\\\ d=\sqrt{64+1}\implies \boxed{d=\sqrt{65}} \\\\[-0.35em] ~\dotfill

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