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February 15, 2020

AHSME 1970 - Probability - Problem 31

What is Probability?


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.

Try the problem

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

Knowledge Graph


AHSME 1970 Problem-knowledge graph

Use some hints


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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