MathematicsCore20 min read

Probability of Combined Events

Adding, multiplying, and knowing which of the two the question wants

This topic appears in:

10th Class Mathematics · ch.12O Level Mathematics (4024) · ch.8A Level Mathematics (9709) · ch.5
01

One decision decides the whole question

Class 9 probability dealt with a single event. This chapter deals with two or more, and almost every mark turns on one decision: are you being asked for this or that, or for this and then that?

OR means add. AND means multiply. Everything else in the chapter is a correction to one of those two rules for the case where the events overlap or interfere with each other.

P(A ∪ B) = P(A) + P(B) − P(A ∩ B)(general addition rule)P(A ∪ B) = P(A) + P(B)(mutually exclusive)P(A ∩ B) = P(A) × P(B)(independent)P(A ∩ B) = P(A) × P(B | A)(dependent)the two special forms are shortcuts; the general forms above them always work
02

Mutually exclusive is not the same as independent

These two words are confused more than any other pair in the syllabus, and they describe completely different situations.

Mutually exclusive events cannot both happen. Drawing a card that is a king and drawing one that is a queen — one card cannot be both, so P(A ∩ B) = 0, and the addition rule loses its correction term.

Independent events can both happen; the point is that one occurring does not change the probability of the other. Two coin tosses, or two draws from a bag with replacement.

Mutually exclusiveIndependent
Can both occur?noyes
P(A ∩ B)0P(A) × P(B)
P(A ∪ B)P(A) + P(B)P(A) + P(B) − P(A)P(B)
Typical exampleking or queen from one cardtwo separate coin tosses

They are almost opposites

If two events with non-zero probabilities are mutually exclusive, they cannot be independent: knowing that A happened tells you B definitely did not, which is a very large change in the probability of B. Any answer describing the same pair of events as both mutually exclusive and independent is wrong.

03

Tree diagrams, and what replacement changes

A tree diagram is the safest tool for any experiment in stages, because it forces you to write down every path. Multiply along a branch to get the probability of one complete outcome; add the branches that satisfy the question.

Two checks are built in. The probabilities on the branches leaving any single node must add to 1. And the probabilities of all the complete outcomes at the far right must also add to 1. If either fails, you have made an error before doing any of the real work.

Worked example

A bag has 5 red and 3 green balls. Two are drawn without replacement. Find P(one of each colour).

  1. First draw: P(R) = 5/8, P(G) = 3/8. Check: they add to 1 ✓Eight balls, five of them red.
  2. Second draw after a red: 7 balls remain, 4 of them red. P(R|R) = 4/7, P(G|R) = 3/7.Both the numerator and the denominator drop for the colour already taken.
  3. Second draw after a green: P(R|G) = 5/7, P(G|G) = 2/7.The reds are untouched, so only the denominator and the green count change.
  4. Two paths give different colours: RG = (5/8)(3/7) = 15/56 and GR = (3/8)(5/7) = 15/56.Both orders count. Using only one is the commonest error in the chapter.
  5. Add them: 15/56 + 15/56 = 30/56 = 15/28.Multiply along, add across — the two rules of the tree.

15/28

Switch between the two modes. The first-stage branches never change — the bag is untouched at that point. Only the second stage differs, because without replacement there is one ball fewer and possibly one fewer of the colour just drawn.

04

Conditional probability

P(B | A) means "the probability of B given that A has already happened". It is what the second stage of a tree diagram is made of, and without replacement every second-stage branch is a conditional probability.

The general multiplication rule P(A ∩ B) = P(A) × P(B|A) is the honest version. When the events happen to be independent, P(B|A) is simply P(B) and the rule collapses to the familiar one — so there is really only one multiplication rule, with a shortcut for the easy case.

Reading "given that" out of the wording

Questions rarely use the notation. They say things like "a student is chosen from those who passed" or "the first ball was red; find the probability that the second is also red". Both are asking for a conditional probability, and the phrase to look for is anything that restricts the sample space before the event of interest.

05

At least one

Whenever the words "at least one" appear, stop and consider the complement. Counting the ways to get at least one of something usually means adding several cases; counting the one way to get none of it is a single multiplication.

For three coins: P(at least one head) as a direct count needs seven of the eight outcomes. Through the complement it is 1 − (1/2)³ = 7/8 in one line.

P(at least one) = 1 − P(none)the single most useful shortcut in the whole chapter

Before you leave this chapter

  1. OR → add, AND → multiply. Decide which before writing anything.
  2. Mutually exclusive means P(A ∩ B) = 0. Independent means P(B|A) = P(B). They are different ideas and cannot both apply.
  3. On a tree, multiply along the branches and add the paths — and check every node sums to 1.
  4. Without replacement, both the numerator and denominator change for the item already taken.
  5. "At least one" is nearly always 1 − P(none).

Practice questions

6 questions · 20 marks · full working on every one

Try each one on paper first, then open the working. The marks are shown where they are actually awarded, because that is where they are actually lost.

Short questions

3 · 6 marks

Two marks each, in the style of the short-question section of the paper. Answer in two or three lines.

SQ1[2 marks]
Distinguish between mutually exclusive and independent events.
Model answer

Mutually exclusive events cannot occur together, so P(A ∩ B) = 0. Independent events can occur together, but one occurring does not affect the probability of the other, so P(A ∩ B) = P(A) × P(B).

Examiner tip. Give the symbolic condition for each. The two definitions in words alone are easy to blur; the symbols are unambiguous.

SQ2[2 marks]
A fair coin is tossed 6 times and lands heads every time. What is the probability of heads on the seventh toss? Explain.
Model answer

1/2. Successive tosses are independent, so the coin has no memory of previous results and the probability is unchanged. Expecting a tail to "even things out" is the gambler's fallacy.

Examiner tip. The reason is the mark. Simply writing 1/2 with no explanation usually scores one out of two.

SQ3[2 marks]
Two balls are drawn from a bag without replacement. Are the two draws independent? Explain.
Model answer

No. Removing the first ball changes both the number of balls left and the composition of the bag, so the probability for the second draw depends on what the first was. The draws are dependent, and the second-stage probabilities are conditional.

Examiner tip. Point to what physically changes — the number and the composition. A bare "no" earns nothing.

Solved numericals

2 · 8 marks

Full working, one step per line, with the marks shown where they are awarded.

N1[4 marks]
A card is drawn from a standard pack of 52. Find the probability that it is a heart or a face card (jack, queen or king).
Full working
  1. P(heart) = 13/52 and P(face) = 12/52twelve face cards: three in each of four suits[1]
  2. The events are not mutually exclusive; there are 3 cards that are bothjack, queen and king of hearts[1]
  3. Uses P(A ∪ B) = P(A) + P(B) − P(A ∩ B) = 13/52 + 12/52 − 3/52[1]
  4. = 22/52 = 11/26[1]

11/26

Examiner tip. Always check for overlap before adding. Adding 13 and 12 straight off gives 25/52, which counts the three heart face cards twice.

N2[4 marks]
The probability that Aisha passes an exam is 0.8 and that Bilal passes is 0.6. The results are independent. Find the probability that (i) both pass, (ii) at least one passes.
Full working
  1. Independent, so P(both) = 0.8 × 0.6[1]
  2. = 0.48[1]
  3. P(neither) = 0.2 × 0.4 = 0.08multiply the two failure probabilities[1]
  4. P(at least one) = 1 − 0.08 = 0.92accept the direct route 0.48 + 0.32 + 0.12 = 0.92[1]

(i) 0.48 (ii) 0.92

Examiner tip. The complement route for part (ii) is one multiplication and one subtraction; the direct route needs three products added. Both are correct, but only one is quick under time pressure.

Long questions

1 · 6 marks

Theory and numerical together, as they appear in the long-question section.

LQ1[6 marks]
A box contains 4 defective and 6 good bulbs. Two bulbs are chosen at random without replacement.
  1. Draw a tree diagram for the two selections.
  2. Find the probability that both bulbs are good.
  3. Find the probability that at least one bulb is defective.
Mark scheme
  1. First stage: P(good) = 6/10, P(defective) = 4/10the two branches must sum to 1[1]
  2. Second stage after a good bulb: 5/9 good, 4/9 defective; after a defective: 6/9 good, 3/9 defectivedenominator 9 throughout, since one bulb has gone[1]
  3. P(both good) = (6/10) × (5/9)multiply along the branch[1]
  4. = 30/90 = 1/3[1]
  5. Uses the complement: P(at least one defective) = 1 − P(both good)"at least one defective" is the exact opposite of "both good"[1]
  6. = 1 − 1/3 = 2/3[1]

(b) 1/3 (c) 2/3

Examiner tip. Part (c) is the complement of part (b), which the examiner has arranged deliberately. Spotting that turns a three-branch calculation into a single subtraction.