Independent and dependent events in probability

Probability questions often ask whether one event affects another. Two shoppers choosing products from the same shelf, for example, may have independent outcomes if the first item is returned, but dependent outcomes if it is removed. Recognising this difference determines which multiplication rule to use.

The distinction is not about whether events happen together. It is about what happens to our expectations after one result is known. Working through conditional probabilities, everyday situations and a raffle example makes that distinction easier to see.

What independence means

An event (A) is independent of an event (B) when learning that (A) occurred gives no new information about (B). For independent events, (P(B\mid A)=P(B)), and the probability that both occur is found by multiplying their separate probabilities:

[ P(A\cap B)=P(A)\times P(B) ]

Two fair coin tosses provide a simple example. Whether the first toss lands heads does not change the chance that the second toss lands heads. Each toss still has a probability of (1/2), so the probability of two heads is (1/2\times1/2=1/4).

Independence also does not mean the events are mutually exclusive. Two coin tosses can both produce heads, even though heads and tails on the same toss are mutually exclusive. For students who want to strengthen the algebra behind bracketed expressions, CurryPlum’s guide to multiplying polynomials shows how products expand and simplify.

When events are dependent

A dependent event changes the probability of another event. Here, the multiplication rule still applies, but the second probability must be conditional:

[ P(A\cap B)=P(A)\times P(B\mid A) ]

For example, imagine a Woolworths display containing 12 chocolate bars, including three Freddo. A customer selects one bar for a prize and sets it aside. If a second bar is selected without replacing the first, the probability that both selections are Freddo is

[ \frac{3}{12}\times\frac{2}{11}=\frac{1}{22} ]

The second fraction is (2/11), rather than (3/12), because the first Freddo is no longer available.

If the first bar were returned to the display before the second selection, the events would be independent. The probability would then be

[ \frac{3}{12}\times\frac{3}{12}=\frac{1}{16} ]

The physical action of replacing the item is what restores the original selection conditions.

Probability in Australian settings

Cricket at the Sydney Cricket Ground shows how changing conditions affect probability. The chance of a shortened match is different when rain arrives before the toss than it is on a dry day. Weather information is therefore conditional evidence, summarised by comparing (P(\text{shortened match}\mid\text{rain})) with (P(\text{shortened match})).

Melbourne provides another familiar situation at Flinders Street Station. A signal fault can change the probability that passengers wait more than ten minutes for the next tram. Without a fault, the usual waiting time applies; with a fault, the conditional probability changes. Similar reasoning applies to service updates across the tram network.

At Adelaide Central Market, a customer chooses one apple from a tray of eight, two of which are bruised. If the chosen apple is removed and another is selected, the second choice is dependent on the first. The probability that the second apple is also bruised becomes (1/7), rather than the original (2/8).

A practical comparison checklist

Identify the connection between the events before calculating. The following checks help separate the two cases without relying on guesswork.

Signs the events are independent

Signs dependence is present

A general study revision checklist can be adapted to record these distinctions alongside formulas, worked examples and common errors. Keeping the condition “with replacement” or “without replacement” visible is especially useful during exam preparation.

Worked example and next step

Consider a raffle at a Sydney school fete, where 40 tickets contain five winning numbers. The organiser draws a ticket, records the winner and returns it to the bowl before drawing again. Because the ticket is replaced, the two draws are independent. The probability of two winning tickets is

[ \frac{5}{40}\times\frac{5}{40}=\frac{1}{64} ]

If winning tickets were kept outside the bowl, the draws would be dependent. The probability would use (5/40) followed by (4/39), giving

[ \frac{5}{40}\times\frac{4}{39}=\frac{1}{78} ]

The concrete next step is to copy this raffle into your notes, create two rows labelled “with replacement” and “without replacement”, and record the matching multiplication rule beside each row.