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What-is-the-probability-that-in-a-class-of-18-people-there-exists-exactly-a-group-of-exactly-3-people-born-on-the-same-day-of-the-week-




Question Number 199353 by mr W last updated on 01/Nov/23
What is the probability that in a class   of 18 people, there exists exactly a   group of exactly 3 people born on the  same day of the week?
$${What}\:{is}\:{the}\:{probability}\:{that}\:{in}\:{a}\:{class}\: \\ $$$${of}\:\mathrm{18}\:{people},\:{there}\:{exists}\:{exactly}\:{a}\: \\ $$$${group}\:{of}\:{exactly}\:\mathrm{3}\:{people}\:{born}\:{on}\:{the} \\ $$$${same}\:{day}\:{of}\:{the}\:{week}? \\ $$
Commented by AST last updated on 01/Nov/23
((C_3 ^(18) ×6^(15) )/7^(18) )≈0.23561
$$\frac{{C}_{\mathrm{3}} ^{\mathrm{18}} ×\mathrm{6}^{\mathrm{15}} }{\mathrm{7}^{\mathrm{18}} }\approx\mathrm{0}.\mathrm{23561} \\ $$
Commented by nikif99 last updated on 03/Nov/23
My approach:  From theory: distributions of n  elements in c cells (with 0 elements  allowed) =C_n ^(n+c−1) =C_(18) ^(18+7−1) =134596  From programming: cases of only one  group of only 3 persons=51646  (cases of one or more groups of only  3 persons=77105).  ⇒ p%=51646/134596=38.37%
$${My}\:{approach}: \\ $$$${From}\:{theory}:\:{distributions}\:{of}\:{n} \\ $$$${elements}\:{in}\:{c}\:{cells}\:\left({with}\:\mathrm{0}\:{elements}\right. \\ $$$$\left.{allowed}\right)\:={C}_{{n}} ^{{n}+{c}−\mathrm{1}} ={C}_{\mathrm{18}} ^{\mathrm{18}+\mathrm{7}−\mathrm{1}} =\mathrm{134596} \\ $$$${From}\:{programming}:\:{cases}\:{of}\:{only}\:{one} \\ $$$${group}\:{of}\:{only}\:\mathrm{3}\:{persons}=\mathrm{51646} \\ $$$$\left({cases}\:{of}\:{one}\:{or}\:{more}\:{groups}\:{of}\:{only}\right. \\ $$$$\left.\mathrm{3}\:{persons}=\mathrm{77105}\right). \\ $$$$\Rightarrow\:{p\%}=\mathrm{51646}/\mathrm{134596}=\mathrm{38}.\mathrm{37\%} \\ $$
Commented by AST last updated on 03/Nov/23
The total number of distribution of 18 people  into 7 days is 7^(18) .  Each person can be born in one of 7 days  So,7×7×7×7...×7(18times)=7^(18)
$${The}\:{total}\:{number}\:{of}\:{distribution}\:{of}\:\mathrm{18}\:{people} \\ $$$${into}\:\mathrm{7}\:{days}\:{is}\:\mathrm{7}^{\mathrm{18}} . \\ $$$${Each}\:{person}\:{can}\:{be}\:{born}\:{in}\:{one}\:{of}\:\mathrm{7}\:{days} \\ $$$${So},\mathrm{7}×\mathrm{7}×\mathrm{7}×\mathrm{7}…×\mathrm{7}\left(\mathrm{18}{times}\right)=\mathrm{7}^{\mathrm{18}} \\ $$
Commented by AST last updated on 03/Nov/23
So,if your program were to be right,the correct  answer may be ((51646)/7^(18) )
$${So},{if}\:{your}\:{program}\:{were}\:{to}\:{be}\:{right},{the}\:{correct} \\ $$$${answer}\:{may}\:{be}\:\frac{\mathrm{51646}}{\mathrm{7}^{\mathrm{18}} } \\ $$
Commented by nikif99 last updated on 03/Nov/23
You are right. The formula I used  refers to non distinctif elements.  For distinctif elements is c^n .
$${You}\:{are}\:{right}.\:{The}\:{formula}\:{I}\:{used} \\ $$$${refers}\:{to}\:{non}\:{distinctif}\:{elements}. \\ $$$${For}\:{distinctif}\:{elements}\:{is}\:{c}^{{n}} . \\ $$

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