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Independent events

Probability · level 4, lesson 4 of 6

A question to start

You toss a coin and roll a die. The chance of heads is \(\tfrac12\). The chance of a 6 is \(\tfrac16\). What is the chance of getting both, heads and a 6?

Someone says: "Add them: \(\tfrac12 + \tfrac16 = \tfrac23\)."

Before working anything out, there is a quick reason why that must be wrong. Getting heads and a 6 is harder than getting heads alone: every time you get both, you certainly got heads, but lots of heads come with a 1, 2, 3, 4 or 5. So the chance of both must be smaller than \(\tfrac12\), not bigger. Adding made it bigger. The right answer is \(\tfrac{1}{12}\), and this lesson shows the rule that gives it, and when you are allowed to use it.

The idea

What "independent" means

Two events are independent if one happening makes no difference to the chance of the other.

  • The coin and the die: the coin has no way of knowing what the die did. Independent.
  • Two rolls of the same die: the die does not remember its last roll. Independent.
  • Two dice rolled together: they bounce off each other, but neither face is pushed towards a particular number. Independent.
  • Your die in Toronto and a friend's die in Vancouver. Independent.

And some that are not independent (they are called dependent):

  • Taking a marble from a bag and not putting it back, then taking another. The first draw changes what is left in the bag, so it changes the chances for the second.
  • "It is cloudy" and "it rains" on the same day. Knowing it is cloudy makes rain more likely.
  • "A student is tall" and "a student is in grade 8" in a school. Older students tend to be taller.

Independence is about the world, not about the numbers. You decide it by asking: "If I knew how the first one turned out, would I change my chances for the second?" If not, they are independent.

The multiplication rule

For independent events \(A\) and \(B\):

\[P(A \text{ and } B) = P(A) \times P(B)\]

For the coin and die: \(P(H \text{ and } 6) = \tfrac12 \times \tfrac16 = \tfrac{1}{12}\).

Check it against the table from lesson 1: there are \(2 \times 6 = 12\) equally likely squares, and exactly one of them is \(H6\). ✓

Why multiply? A picture

Draw a square and call its area 1: it stands for "everything that can happen". Split it into two halves, left for heads and right for tails. Then split it into six rows, one for each face of the die.

1 2 3 4 5 6 heads (1/2) tails (1/2) H6 the whole square = 1 green = heads and a 6 = 1/6 of 1/2 = 1/12

Because the coin and the die are independent, the die's six rows run straight across both halves, in the same proportions. The heads half is split into sixths exactly like the tails half. (If the die "preferred" 6 after heads, the rows would be different sizes on the two sides.)

The green piece, heads and 6, is one row of the left half. It is \(\tfrac16\) of \(\tfrac12\) of the square:

\[\frac16 \text{ of } \frac12 = \frac16 \times \frac12 = \frac{1}{12}\]

"Of" means multiply, just as \(\tfrac13\) of 12 is \(\tfrac13 \times 12 = 4\). So the multiplication rule is really just saying: the second event happens in the same fraction of the first event's cases as it does overall.

The same picture explains why the answer is smaller than both \(\tfrac12\) and \(\tfrac16\). Multiplying by a fraction less than 1 makes things smaller: a part of a part is smaller than the whole part.

Multiply along the branches

In lesson 2 you multiplied along the branches of a tree. That was this rule at work: coin flips are independent, so the chance of \(H\) then \(H\) then \(T\) is \(\tfrac12 \times \tfrac12 \times \tfrac12\).

Here is the tree for rolling two dice and looking for sixes.

1/6 5/6 1/6 5/6 1/6 5/6 six not 6 six not 6 six not 6 six, six: 1/36 six, not: 5/36 not, six: 5/36 not, not: 25/36 1st die 2nd die

The branches for the second die are the same whichever way the first die went (\(\tfrac16\) and \(\tfrac56\) both times). That is what independence looks like on a tree: the second set of branches does not change.

  • Two sixes: \(\tfrac16 \times \tfrac16 = \tfrac{1}{36}\).
  • No sixes: \(\tfrac56 \times \tfrac56 = \tfrac{25}{36}\).
  • Exactly one six: \(\tfrac{5}{36} + \tfrac{5}{36} = \tfrac{10}{36} = \tfrac{5}{18}\).

Check: \(\tfrac{1 + 5 + 5 + 25}{36} = 1\). ✓ And it agrees with the 36-square table in level 3: \(1 + 10 = 11\) squares with at least one 6.

More than two

The rule keeps going. For independent events, multiply all their probabilities. Five coins all heads:

\[\frac12 \times \frac12 \times \frac12 \times \frac12 \times \frac12 = \frac{1}{32}\]

Each extra coin halves the chance. Ten coins all heads is \(\tfrac{1}{1024}\), a little less than one in a thousand.

When you may not multiply like this

If the events are dependent, the chances for the second event change, and multiplying the two "starting" chances gives the wrong answer. Take a bag with 3 red and 2 blue marbles.

  • With putting back: each draw is red with probability \(\tfrac35\). Two reds: \(\tfrac35 \times \tfrac35 = \tfrac{9}{25}\).
  • Without putting back: after a red is taken out, the bag has 2 red and 2 blue. Now the second draw is red with probability \(\tfrac24\), not \(\tfrac35\). The draws are dependent.

Level 5 shows how trees handle that case: you still multiply along a branch, but you use the changed chances.

Worked examples

1. Heads and even. A coin and a die. \(P(H) = \tfrac12\) and \(P(\text{even}) = \tfrac12\), and they are independent. \(P(H \text{ and even}) = \tfrac12 \times \tfrac12 = \tfrac14\). Lesson 1 found \(\tfrac{3}{12} = \tfrac14\) by counting squares. ✓

2. Three dice, all even. Each die is even with probability \(\tfrac12\), independently: \(\tfrac12 \times \tfrac12 \times \tfrac12 = \tfrac18\). (Listing would need all \(6 \times 6 \times 6 = 216\) outcomes; there are \(3 \times 3 \times 3 = 27\) all-even ones, and \(\tfrac{27}{216} = \tfrac18\). ✓)

3. A spinner and a die. A spinner is red with probability \(\tfrac13\). It is spun, and a die is rolled. Find \(P(\text{red and more than } 4)\).

\(P(\text{more than } 4) = \tfrac26 = \tfrac13\). Independent, so \(\tfrac13 \times \tfrac13 = \tfrac19\).

4. Decimals work too. A basketball player scores a free throw with probability 0.8, and we assume each shot is independent of the last.

  • Two in a row: \(0.8 \times 0.8 = 0.64\).
  • Miss both: \(0.2 \times 0.2 = 0.04\).
  • At least one: \(1 - 0.04 = 0.96\).
  • Exactly one: \(0.8 \times 0.2 + 0.2 \times 0.8 = 0.32\). Check: \(0.64 + 0.32 + 0.04 = 1\). ✓

("We assume" matters. A real player might get nervous after a miss, which would make the shots dependent. Saying "independent" is a choice about how to model the world, and it should be a sensible one.)

5. All different on three dice. Roll three dice. What is the chance they all show different numbers?

  • The first die can be anything.
  • The second must differ from the first: \(\tfrac56\).
  • The third must differ from both: \(\tfrac46\).

\(1 \times \tfrac56 \times \tfrac46 = \tfrac{20}{36} = \tfrac59\). The dice are independent; the chances \(\tfrac56\) and \(\tfrac46\) come from what we need, not from the dice changing.

Try it

  1. A coin is tossed and a die rolled. Find \(P(H \text{ and } 3)\) by multiplying, and check with the 12-square table.
  2. Two dice are rolled. Find \(P(\text{both odd})\) and \(P(\text{first is } 6 \text{ and second is not } 6)\).
  3. Four coins are tossed. Find \(P(\text{all tails})\) and \(P(\text{at least one head})\).
  4. A spinner wins with probability 0.3, independently each spin. It is spun twice. Find \(P(\text{win both})\), \(P(\text{win neither})\) and \(P(\text{win exactly one})\). Check that they add to 1.
  5. Independent or not? (a) two flips of one coin; (b) "it is cloudy" and "it rains" today; (c) the first and second cards drawn from a pack without putting the first back; (d) your die roll and your friend's in another city.
  6. Draw an area picture (a square split into parts) for a coin and a spinner that is red with probability \(\tfrac14\). Shade "heads and red" and find its probability.
  7. Mo says: "The chance of a 6 is \(\tfrac16\), so the chance of two 6s is \(\tfrac16 + \tfrac16 = \tfrac13\)." Give two reasons why that cannot be right.
  8. A die is rolled 3 times. Find \(P(\text{no } 6 \text{ at all})\) and \(P(\text{at least one } 6)\), as fractions and as percents (to one decimal place).
Answers
  1. \(\tfrac12 \times \tfrac16 = \tfrac{1}{12}\). In the table, only the square \(H3\): \(\tfrac{1}{12}\). ✓
  2. Both odd: \(\tfrac12 \times \tfrac12 = \tfrac14\). First 6, second not: \(\tfrac16 \times \tfrac56 = \tfrac{5}{36}\).
  3. All tails: \(\left(\tfrac12\right)^4 = \tfrac{1}{16}\). At least one head: \(1 - \tfrac{1}{16} = \tfrac{15}{16}\).
  4. Both: \(0.3 \times 0.3 = 0.09\). Neither: \(0.7 \times 0.7 = 0.49\). Exactly one: \(0.3 \times 0.7 + 0.7 \times 0.3 = 0.42\). Sum: \(0.09 + 0.49 + 0.42 = 1\). ✓
  5. (a) Independent. (b) Dependent: clouds make rain more likely. (c) Dependent: the first card is no longer in the pack. (d) Independent.
  6. Split the square down the middle (heads, tails) and across into 4 equal rows (one row red, three not). "Heads and red" is one row of the left half: \(\tfrac14 \times \tfrac12 = \tfrac18\) of the square.
  7. First, "two 6s" must be less likely than one 6, but \(\tfrac13\) is more than \(\tfrac16\). Second, by the same thinking, six 6s would have chance \(\tfrac66 = 1\), certain, which is silly. The right answer is \(\tfrac16 \times \tfrac16 = \tfrac{1}{36}\).
  8. No 6: \(\tfrac56 \times \tfrac56 \times \tfrac56 = \tfrac{125}{216}\), about \(0.579\), or \(57.9\%\). At least one 6: \(1 - \tfrac{125}{216} = \tfrac{91}{216}\), about \(0.421\), or \(42.1\%\).

Common mistake

Adding for "and". Adding is for "or" (and only when the events cannot both happen). For "and" with independent events, multiply. A quick sense check: "A and B" can never be more likely than A alone.

Multiplying when the events are dependent. If the first result changes the chances for the second (no putting back, weather on the same day), the plain multiplication rule does not apply.

Where it leads