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CMS Math Olympiad · 6th Grade

Probability

What is the probability of rolling a six? Of pulling out a red ball? Let’s learn to calculate chances like mathematicians.

1
What is probability?

Probability is a number from 0 to 1 that shows how likely something is to happen.

Formula
\(P = \dfrac{\text{number of favourable outcomes}}{\text{total number of outcomes}}\)
What the extreme values mean
\(P = 0\) — the event is impossible (rolling a 7 on an ordinary die).
\(P = 1\) — the event is certain (you will roll a number from 1 to 6).
\(P = \dfrac{1}{2}\) — equally likely to happen or not.
Example — coin toss

We toss a coin. What is the probability of getting heads?

Total outcomes: 2 (heads or tails).

Favourable outcomes: 1 (heads).

\(P = \dfrac{1}{2}\).

Try it
A bag contains 3 red and 7 blue balls. The probability of drawing a red ball is:
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2
Rolling a die

An ordinary die has 6 faces with numbers 1, 2, 3, 4, 5, 6. Each number is equally likely.

1 2 3 4 5 6 Each has probability 1/6
Typical questions

Probability of an even number: 2, 4, 6 → \(P = \dfrac{3}{6} = \dfrac{1}{2}\).

Probability of rolling a number greater than 4: numbers 5 and 6 → \(P = \dfrac{2}{6} = \dfrac{1}{3}\).

Probability of a prime number: 2, 3, 5 → \(P = \dfrac{3}{6} = \dfrac{1}{2}\).

Try it
We roll a die. The probability that the result is divisible by 3 is:
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3
Balls in a bag

This is the most popular CMS type. Usually there is a bag with balls of different colours, and you need the probability of drawing a ball of the required colour.

3 red 5 blue
Example

There are 3 red and 5 blue balls in a bag. We draw one at random.

Probability of red: \(\dfrac{3}{3+5} = \dfrac{3}{8}\).

Probability of blue: \(\dfrac{5}{8}\).

Careful: without replacement
If you drew a ball and did not put it did not return it to the bag, the total number decreases!

Originally: 3 red + 5 blue = 8. After drawing one red → 2 red + 5 blue remain = 7.
Probability of a second red: \(\dfrac{2}{7}\), not \(\dfrac{2}{8}\)!
Try it
A bag contains 4 green, 3 yellow, and 5 white balls. The probability of drawing a yellow ball is:
Try it
A bag contains 6 red and 4 white balls. One red ball is drawn and not replaced. The probability that the second one is also red is:
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4
Probability of “NOT”
Complementary event
\(P(\text{NOT } A) = 1 - P(A)\)
The probability that an event does NOT happen equals 1 minus the probability that it does happen.

This is a very powerful trick! Sometimes it is easier to calculate what we do not wantand subtract from 1.

Example

We roll a die. What is the probability of NOT getting a six?

\(P(\text{six}) = \dfrac{1}{6}\).

\(P(\text{not six}) = 1 - \dfrac{1}{6} = \dfrac{5}{6}\).

Try it
A bag contains 2 red and 8 blue balls. The probability of drawing NOT a red ball is:
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5
Two events: “AND” and “OR”
“AND” — both together
We toss a coin twice. What is the probability that both times get heads?

First toss: \(P = \dfrac{1}{2}\). Second toss: \(P = \dfrac{1}{2}\).
Together: \(\dfrac{1}{2} \times \dfrac{1}{2} = \dfrac{1}{4}\).

Rule: if events are independent, we multiply their probabilities.
“OR” — at least one
We roll a die. Probability of getting 2 or 5?

These are two different outcomes that do not overlap:
\(P = \dfrac{1}{6} + \dfrac{1}{6} = \dfrac{2}{6} = \dfrac{1}{3}\).

Rule: if events cannot happen at the same time, we add the probabilities.
How to remember it
“AND” → multiply (you need both to happen — this is rarer).
“OR” → add (it is enough that at least one happens — this is more common).
Try it
We roll two dice. The probability that both show six is:
Try it
We roll a die. The probability of getting 1, 3, or 5 (an odd number) is:
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6
Olympiad practice
Problem 1 · Easy
A box contains 5 red, 3 green, and 2 yellow pencils. One is chosen at random. The probability it is green is:
Problem 2 · Easy
A number from 1 to 20 is chosen at random. The probability that it is even is:
Problem 3 · Medium
We roll two dice and add the results. The probability that the sum is 7 is:
Total outcomes for two dice: \(6 \times 6 = 36\).

Pairs summing to 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) → 6 pairs.

\(P = \dfrac{6}{36} = \dfrac{1}{6}\).
Problem 4 · Medium
A number from 1 to 30 is chosen at random. The probability that it is divisible by 5 is:
Numbers from 1 to 30 divisible by 5: 5, 10, 15, 20, 25, 30 → 6 numbers.
\(P = \dfrac{6}{30} = \dfrac{1}{5}\).

(Take note: options A and C are the same number! \(\dfrac{5}{30} = \dfrac{1}{6}\), not \(\dfrac{1}{5}\). In CMS always simplify fractions.)
Problem 5 · Medium
A bag contains 4 white and 6 black balls. Two are drawn at random without replacement. The probability that both are black is:
First black: \(\dfrac{6}{10}\).
After that, 5 black balls remain out of 9 → second black: \(\dfrac{5}{9}\).

Both: \(\dfrac{6}{10} \times \dfrac{5}{9} = \dfrac{30}{90} = \dfrac{1}{3}\).
Problem 6 · Medium+
Three numbers are chosen at random from the set \(\{1, 2, 3, 4, 5, 6\}\). How many ways are there to choose three numbers with an even sum?
Even: {2, 4, 6}. Odd: {1, 3, 5}.

The sum is even if all three are even, or one is even and two are odd.

All three even: \(\binom{3}{3} = 1\) way.
One even + two odd: \(\binom{3}{1} \times \binom{3}{2} = 3 \times 3 = 9\) ways.

Total: \(1 + 9 = 10\).

(This is CMS 2021 Q24!)
Cheat sheet — Probability
Basic formula: \(P = \dfrac{\text{favourable}}{\text{total}}\)
Complementary event: \(P(\text{NOT }A) = 1 - P(A)\)
“AND” (both): multiply the probabilities
“OR” (at least one): add the probabilities
Without replacement: after each draw, reduce both the numerator and the denominator
Die: 6 outcomes, each with probability \(\dfrac{1}{6}\)
Two dice: 36 outcomes

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