Cambridge International AS & A Level Mathematics 9709 (2026-2027) · Paper 5: Probability & Statistics 1 · official syllabus
PROBABILITY LAB
Learn it by playing it
Four short missions on equally likely outcomes, overlapping events and sampling without replacement. Predict the rule, then calculate. Keep a joint event distinct from a condition.
1234
Mission 1 of 4
Count the equally likely outcomes
Two fair independent six-sided dice are thrown. Each ordered pair is equally likely, so there are 36 outcomes. Set the number of ordered pairs giving a total of 8. The grid highlights this event; divide your count by 36 for its probability.
The possible dice totals from 2 to 12 are...
Each of the 36 cells is one equally likely ordered outcome. Teal cells give total 8; rows and columns distinguish the two dice.
Your count is 1; count ordered pairs, not the number of possible totals.
Study notes
Played the lab above? These notes are the reference version of what you just felt - use them to check exact formulas and the exam wording.
Define outcomes before counting 5.3
A probability is favourable outcomes divided by all outcomes only when those elementary outcomes are equally likely. State the experiment and sample space. For two fair independent dice, ordered pairs (first die, second die) form 36 equally likely outcomes; the possible totals 2 to 12 are not equally likely.
Worked example: a total of 8 has the outcomes (2,6), (3,5), (4,4), (5,3), (6,2), so its probability is 5/36. Do not use 1/11 merely because there are eleven possible totals. Counting the two dice as unordered pairs also needs unequal weights, so the ordered sample space is simpler.
Use permutations or combinations when they count equally likely outcomes for the specified sampling method. Whether order matters for the count is not the same question as whether the desired event has been stated in words that mention order.
Complements and mutually exclusive cases 5.3
The complement A' contains outcomes not in A, so P(A')=1-P(A). Mutually exclusive events cannot happen on the same trial; their intersection has probability zero. Add probabilities for disjoint cases that cover the event required.
For two events that overlap, adding P(A)+P(B) counts their intersection twice. Subtract it once to find P(A or B). A diagram or an outcome list can make the overlap clear. 'Or' in a probability question normally includes both unless the question says exactly one.
Worked example: on a fair die let A be an even score and B a score above 4. A={2,4,6}, B={5,6}, and their intersection is {6}. Therefore P(A or B)=3/6+2/6-1/6=4/6=2/3. These events are not mutually exclusive.
Follow a probability tree one branch at a time 5.3
Multiply along a path using the relevant conditional probability at each stage. Add the resulting path probabilities for disjoint ways to reach the event. Branch probabilities leaving the same node sum to 1.
A bag contains 3 red and 2 blue balls. Two balls are drawn at random without replacement. P(first red, second blue)=(3/5)(2/4)=3/10. Once a red is drawn, four balls remain and two are blue. Keeping the second denominator at 5 would model replacement instead.
For one red and one blue in either order, add the disjoint paths RB and BR: (3/5)(2/4)+(2/5)(3/4)=3/5. The same event can be counted by combinations: (3C1)(2C1)/(5C2)=6/10=3/5. These two routes agree because they describe the same experiment and event.
Conditional probability restricts the sample space 5.3
P(A|B) means the probability of A given that B has happened. For P(B)>0, P(A|B)=P(A and B)/P(B). The denominator is the conditioning event, not automatically the probability named first in a sentence.
Worked example: in the same bag experiment, given that the first ball was red, the probability that the second is blue is 2/4=1/2. This differs from the joint probability 3/10 of first red and second blue. The condition has already fixed the first draw.
Reverse conditions need a new calculation. Given that exactly one red and one blue were drawn, the probability that the first was red is P(RB)/P(one of each)=(3/10)/(3/5)=1/2. Do not reverse a conditional by simply swapping the letters.
Independence is not the same as exclusion 5.3
Events A and B are independent when P(A and B)=P(A)P(B). If P(B)>0, this is equivalent to P(A|B)=P(A). Independence means learning B does not change A's probability. It must follow from the experiment or a calculation, not from how unrelated the event names sound.
Worked example on a fair die: A={2,4,6} and C={3,6}. Their intersection is {6}, with probability 1/6. Since P(A)P(C)=(1/2)(1/3)=1/6, A and C are independent. They can both occur, so they are not mutually exclusive.
Two mutually exclusive events with positive probabilities cannot be independent: their intersection probability is zero, but the product of their positive probabilities is not zero. Events on separate trials can be independent under an appropriate model; drawing without replacement generally changes later probabilities.
Check the model and the final probability 5.3
Probabilities lie between 0 and 1. A complete distribution or a node's outgoing tree branches must sum to 1. Check whether an event is impossible or certain before doing a long calculation. Keep exact fractions in working and round only at the requested stage.
Clearly distinguish at least one, exactly one and more than one. For two independent trials with success probability p, at least one success has probability 1-(1-p)^2; exactly one has probability 2p(1-p). The first includes two successes; the second does not.
With p=1/4, at least one success in two independent trials is 1-(3/4)^2=7/16; exactly one is 2(1/4)(3/4)=3/8. These formulas rely on the specified independence and common success probability, rather than applying to every two-stage experiment.
EduTribe study notes - original material aligned to the Cambridge International AS & A Level Mathematics 9709 syllabus (2026-2027). Automated checks completed; not yet reviewed by an educator.
Free public preview - no sign-in or pass needed. Automated checks completed; not yet reviewed by an educator. Sign in to save your practice progress across sessions.