# Sample Probability Problems from AMC

The very first time a probability problem has been included in the Annual High School Mathematics Contest happened in 1970. The problem (#31) reads:

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?

(A) 2/5 (B) 1/5 (C) 1/6 (D) 1/11 (E) 1/15

In 1971 there was one problem (#23):

Teams A and B are playing a series of games. If the odds for either team to win any game are even and Team A must win two or Team B three games to win the series, then the odds favoring Team A to win the series are:

(A) 11 to 5 (B) 5 to 2 (C) 8 to 3 (D) 3 to 2 (E) 13 to 6

In 1972 there was also one problem (#17):

A piece of string is cut in two at a point selected at random. The probability that the longer piece is at least x times as large as the shorter piece

(A) 1/2 (B) 2/x (C) 1/(x+1) (D) 1/x (E) 2/(x+1)

Since then no one has been surprised by the presence of a probability problem. For example, #26 (1981)

Alice, Bob, and Carol repeatedly take turns tossing a die. Alice begins; Bob always follows Alice; Carol always follows Bob, and Alice always follows Carol. Find the probability that Carol will be the first one to toss a six. (The probability of obtaining a six on any toss is 1/6, independent of the outcomes of any other toss.)

(A) 1/3 (B) 2/9 (C) 5/18 (D) 25/91 (E) 36/91

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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?

(A) 2/5 (B) 1/5 (C) 1/6 (D) 1/11 (E) 1/15

### Solution

The maximum sum of 5 decimal digits is 45. To obtain 43 we should allow either one 7 or two 8s. There are five integers of the first kind 79999, 97999, 99799, 99979, 99997, and 10 integers of the second kind 88999, 89899, 89989, 89998, 98899, 98989, 98998, 99889, 99898, 99988.

Numbers are divisible by 11 is their alternating sum of digits are divisible by 11. There are two numbers of the first kind that satisfies this criterion: 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.

Teams A and B are playing a series of games. If the odds for either team to win any game are even and Team A must win two or Team B three games to win the series, then the odds favoring Team A to win the series are:

(A) 11 to 5 (B) 5 to 2 (C) 8 to 3 (D) 3 to 2 (E) 13 to 6

### Solution

The series would last at most 4 games before either Team A wins two games or loses (to Team B) three games. Team A wins the series in two games with probability 1/4. There are two ways for it to win in three games: LWW or WLW; both occur with probability 1/8 so that the probability that Team A wins in exactly 3 games is also 1/4. Team A wins in exactly four games is one of the following events takes place: WLLW, LWLW, LLWW. Each of these happens with probability 1/16 to the total of 3/16. The total probability to win the series for Team A is, therefore,

A piece of string is cut in two at a point selected at random. The probability that the longer piece is at least x times as large as the shorter piece

(A) 1/2 (B) 2/x (C) 1/(x+1) (D) 1/x (E) 2/(x+1)

Think of the string as a segment AB of length 1 + x. Let P be a point inside AB such that

Alice, Bob, and Carol repeatedly take turns tossing a die. Alice begins; Bob always follows Alice; Carol always follows Bob, and Alice always follows Carol. Find the probability that Carol will be the first one to toss a six. (The probability of obtaining a six on any toss is 1/6, independent of the outcomes of any other toss.)

(A) 1/3 (B) 2/9 (C) 5/18 (D) 25/91 (E) 36/91

For Carol to win in the first round, she has to come up with a six after both Alice and Bob got anything else, but six. This happens with probability ^{2}/6^{3}.^{3}/6^{3}^{2}/6^{3}. But the second round is reached only after Carol fails to win on the first. The two events being independent, the probability of Alice winning on the second round is the product: 5^{3}/6^{3}×5^{2}/6^{3}. She fails to win on the second round with the probability of ^{3}/6^{3}×5^{3}/6^{3} = 5^{3·2}/6^{3·2}

The evaluations of probabilities continues in this manner. Carol wins on the third round with the probability

5^{3·2}/6^{3·2}×5^{2}/6^{3};

she wins the fourth round with the probability of

5^{3·3}/6^{3·3}×5^{2}/6^{3};

and so forth. Carol wins the competition of she wins on one of the rounds, and this happens with the probability

5^{2}/6^{3} + 5^{3}/6^{3}×5^{2}/6^{3} + (5^{3}/6^{3})^{2}×5^{2}/6^{3} + (5^{3}/6^{3})^{3}×5^{2}/6^{3} + ...

This is the sum of a geometric series, which me know to be equal to

5^{2}/6^{3} × 1 / (1 - 5^{3}/6^{3}) = 5^{2} / (6^{3} - 5^{3}) = 25/91.

Denote the event of a person T being the first to toss a six as W(T). We just found that the probability

Clearly, one of the three participants will be the first to through a six. We can check that

P(W(Bob)) = 5^{1} / 6^{2} × 6^{3} / (6^{3} - 5^{3}) = 30/91

and

P(W(Alice)) = 1 / 6 × 6^{3} / (6^{3} - 5^{3}) = 36/91.

As expected, the sum of the three probabilities is 1: 25/91 + 30/91 + 36/91 = 1. The event that six does not come up in an infinite series of throws is improbable - it's probability equals 0.

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Copyright © 1996-2018 Alexander Bogomolny