Probability is a branch of mathematics that deals with calculating the likelihood of a given event's occurrence, which is expressed as a number between $1$ and $0$. ... Each coin toss is an independent event; the outcome of one trial has no effect on subsequent ones.This Problem has taken from AHSME 1970.
If the number is selected at random from the set of all five-digit numbers in which the sum of the digits is equal to $43$, what is the probability that the number will be divisible by $11$?
$\textbf{(A)} \quad \frac25 \quad \textbf{(B)} \quad \frac15 \quad \textbf{(C)} \quad \frac25 \quad \textbf{(D)} \quad \frac{1}{11} \quad \textbf{(E)} \quad \frac{1}{15}$
AHSME 1970 Problem 31
Probability
3 out of 10
Mathematics Circle
we know the maximum sum of $5$ decimal digits is $45$. To obtain sum of the digits, $43$ we should allow either one $7$ or two \(8\). There are five integers of the first kind $79999, 97999, 99799, 99979, 99997$,one $7$ is here of each integers. And $10$ integers of the second kind $ 88999, 89899, 89989, 89998, 98899, 98989, 98998, 99889, 99898, 99988$, two $8$ are here of each integers.
Numbers are divisible by \(11\). It means their alternating sum of digits are divisible by $11$. There are two numbers of the first kind that satisfies this criterion. They are $97999$ and $99979$. Indeed, $9 - 7 + 9 - 9 + 9 = 11$, and
\(9 - 9 + 9 - 7 + 9 = 11\).
Of the second kind, only one number $98989$ is divisible by $11$. Indeed ,$ 9 - 8 + 9 - 8 + 9 = 11$.
Out of the total of $15$ numbers, three are divisible by $11$. The probability of this event is $3/15 = 1/5$.
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