{"id":104,"date":"2026-08-07T12:40:09","date_gmt":"2026-08-07T12:40:09","guid":{"rendered":"https:\/\/sqoolpapers.co.za\/notes\/probability-g12-maths-t4p1\/"},"modified":"2026-08-22T16:36:47","modified_gmt":"2026-08-22T16:36:47","slug":"probability-g12-maths-t4p1","status":"publish","type":"post","link":"https:\/\/repstaq.com\/study-guide-docs\/probability-g12-maths-t4p1\/","title":{"rendered":"Probability-G12-Maths-T4P1"},"content":{"rendered":"\n

Probability<\/h2>\n\n\n\n

Probability measures how likely an event is to occur. It is used to analyse uncertainty and make predictions. In examinations, show the correct formula, substitute values clearly, simplify accurately and give the final answer. This method-based approach is important because marks are awarded for correct mathematical steps as well as the answer.<\/p>\n\n\n\n

Basic probability<\/h2>\n\n\n\n

An experiment is a process with an uncertain result, such as tossing a coin. An outcome is one possible result, while the sample space is the set of all possible outcomes. An event is a selection of outcomes from the sample space.<\/p>\n\n\n\n

If all outcomes are equally likely, probability is calculated using:<\/p>\n\n\n\n

P<\/mi>(<\/mo>A<\/mi>)<\/mo>=<\/mo>n<\/mi>(<\/mo>A<\/mi>)<\/mo><\/mrow>n<\/mi>(<\/mo>S<\/mi>)<\/mo><\/mrow><\/mfrac><\/mrow>P(A)=\\frac{n(A)}{n(S)}<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

Here, the numerator is the number of outcomes favourable to event A, and the denominator is the total number of outcomes in the sample space.<\/p>\n\n\n\n

Probability always lies between zero and one:<\/p>\n\n\n\n

0<\/mn>\u2264<\/mo>P<\/mi>(<\/mo>A<\/mi>)<\/mo>\u2264<\/mo>1<\/mn><\/mrow>0\\leq P(A)\\leq 1<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

An impossible event has probability zero, while a certain event has probability one.<\/p>\n\n\n\n

Example<\/strong><\/p>\n\n\n\n

A fair six-sided die is rolled. Calculate the probability of obtaining a number greater than 4.<\/p>\n\n\n\n

The sample space is:<\/p>\n\n\n\n

S<\/mi>=<\/mo>{<\/mo>1<\/mn>;<\/mo>2<\/mn>;<\/mo>3<\/mn>;<\/mo>4<\/mn>;<\/mo>5<\/mn>;<\/mo>6<\/mn>}<\/mo><\/mrow>S=\\{1;2;3;4;5;6\\}<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

The favourable outcomes are 5 and 6. Therefore:<\/p>\n\n\n\n

P<\/mi>(<\/mo>number greater than <\/mtext>4<\/mn>)<\/mo>=<\/mo>2<\/mn>6<\/mn><\/mfrac><\/mrow>P(\\text{number greater than }4)=\\frac{2}{6}<\/annotation><\/semantics><\/math><\/div>\n\n\n\n
P<\/mi>(<\/mo>number greater than <\/mtext>4<\/mn>)<\/mo>=<\/mo>1<\/mn>3<\/mn><\/mfrac><\/mrow>P(\\text{number greater than }4)=\\frac{1}{3}<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

For experimental probability, use the results of repeated trials:<\/p>\n\n\n\n

P<\/mi>(<\/mo>A<\/mi>)<\/mo>=<\/mo>number of times <\/mtext>A<\/mi> occurs<\/mtext><\/mrow>total number of trials<\/mtext><\/mfrac><\/mrow>P(A)=\\frac{\\text{number of times }A\\text{ occurs}}{\\text{total number of trials}}<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

Complementary events<\/h2>\n\n\n\n

The complement of event A, written as the event \u201cnot A\u201d, contains all outcomes in the sample space that are not in A.<\/p>\n\n\n\n

P<\/mi>(<\/mo>A<\/mi>\u2032<\/mo><\/msup>)<\/mo>=<\/mo>1<\/mn>\u2212<\/mo>P<\/mi>(<\/mo>A<\/mi>)<\/mo><\/mrow>P(A’)=1-P(A)<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

Equivalently:<\/p>\n\n\n\n

P<\/mi>(<\/mo>A<\/mi>)<\/mo>+<\/mo>P<\/mi>(<\/mo>A<\/mi>\u2032<\/mo><\/msup>)<\/mo>=<\/mo>1<\/mn><\/mrow>P(A)+P(A’)=1<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

Example<\/strong><\/p>\n\n\n\n

The probability that a learner passes a test is 0,82. Calculate the probability that the learner does not pass.<\/p>\n\n\n\n

P<\/mi>(<\/mo>not pass<\/mtext>)<\/mo>=<\/mo>1<\/mn>\u2212<\/mo>P<\/mi>(<\/mo>pass<\/mtext>)<\/mo><\/mrow>P(\\text{not pass})=1-P(\\text{pass})<\/annotation><\/semantics><\/math><\/div>\n\n\n\n
P<\/mi>(<\/mo>not pass<\/mtext>)<\/mo>=<\/mo>1<\/mn>\u2212<\/mo>0,82<\/mn><\/mrow>P(\\text{not pass})=1-0{,}82<\/annotation><\/semantics><\/math><\/div>\n\n\n\n
P<\/mi>(<\/mo>not pass<\/mtext>)<\/mo>=<\/mo>0,18<\/mn><\/mrow>P(\\text{not pass})=0{,}18<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

Exam Tip<\/strong><\/p>\n\n\n\n

Words such as \u201cnot\u201d, \u201cdoes not\u201d, \u201cnone\u201d and \u201cat least one\u201d often indicate that a complementary event may simplify the calculation.<\/p>\n\n\n\n

For example:<\/p>\n\n\n\n

P<\/mi>(<\/mo>at least one success<\/mtext>)<\/mo>=<\/mo>1<\/mn>\u2212<\/mo>P<\/mi>(<\/mo>no successes<\/mtext>)<\/mo><\/mrow>P(\\text{at least one success})=1-P(\\text{no successes})<\/annotation><\/semantics><\/math><\/div>\n\n\n\n
\"\"\/<\/a><\/figure>\n\n\n\n

Mutually exclusive events<\/h2>\n\n\n\n

Two events are mutually exclusive if they cannot occur at the same time. They have no outcomes in common.<\/p>\n\n\n\n

P<\/mi>(<\/mo>A<\/mi>\u2229<\/mo>B<\/mi>)<\/mo>=<\/mo>0<\/mn><\/mrow>P(A\\cap B)=0<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

For mutually exclusive events, the addition rule becomes:<\/p>\n\n\n\n

P<\/mi>(<\/mo>A<\/mi>\u222a<\/mo>B<\/mi>)<\/mo>=<\/mo>P<\/mi>(<\/mo>A<\/mi>)<\/mo>+<\/mo>P<\/mi>(<\/mo>B<\/mi>)<\/mo><\/mrow>P(A\\cup B)=P(A)+P(B)<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

Example<\/strong><\/p>\n\n\n\n

A card numbered from 1 to 10 is selected randomly. Let event A be selecting a number less than 3 and event B be selecting a number greater than 8.<\/p>\n\n\n\n

A<\/mi>=<\/mo>{<\/mo>1<\/mn>;<\/mo>2<\/mn>}<\/mo><\/mrow>A=\\{1;2\\}<\/annotation><\/semantics><\/math><\/div>\n\n\n\n
B<\/mi>=<\/mo>{<\/mo>9<\/mn>;<\/mo>10<\/mn>}<\/mo><\/mrow>B=\\{9;10\\}<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

The events have no common outcomes. Therefore:<\/p>\n\n\n\n

P<\/mi>(<\/mo>A<\/mi>\u222a<\/mo>B<\/mi>)<\/mo>=<\/mo>P<\/mi>(<\/mo>A<\/mi>)<\/mo>+<\/mo>P<\/mi>(<\/mo>B<\/mi>)<\/mo><\/mrow>P(A\\cup B)=P(A)+P(B)<\/annotation><\/semantics><\/math><\/div>\n\n\n\n
P<\/mi>(<\/mo>A<\/mi>\u222a<\/mo>B<\/mi>)<\/mo>=<\/mo>2<\/mn>10<\/mn><\/mfrac>+<\/mo>2<\/mn>10<\/mn><\/mfrac><\/mrow>P(A\\cup B)=\\frac{2}{10}+\\frac{2}{10}<\/annotation><\/semantics><\/math><\/div>\n\n\n\n
P<\/mi>(<\/mo>A<\/mi>\u222a<\/mo>B<\/mi>)<\/mo>=<\/mo>4<\/mn>10<\/mn><\/mfrac>=<\/mo>2<\/mn>5<\/mn><\/mfrac><\/mrow>P(A\\cup B)=\\frac{4}{10}=\\frac{2}{5}<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

Common Mistake<\/strong><\/p>\n\n\n\n

Mutually exclusive events are not the same as independent events. Mutually exclusive events cannot occur together. Independent events can occur together, but one does not affect the probability of the other.<\/p>\n\n\n\n

Independent and dependent events<\/h2>\n\n\n\n

Events A and B are independent if the occurrence of one event does not affect the probability of the other.<\/p>\n\n\n\n

The test for independence is:<\/p>\n\n\n\n

P<\/mi>(<\/mo>A<\/mi>\u2229<\/mo>B<\/mi>)<\/mo>=<\/mo>P<\/mi>(<\/mo>A<\/mi>)<\/mo>\u00d7<\/mo>P<\/mi>(<\/mo>B<\/mi>)<\/mo><\/mrow>P(A\\cap B)=P(A)\\times P(B)<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

Example<\/strong><\/p>\n\n\n\n

A fair coin is tossed and a fair die is rolled. Calculate the probability of obtaining heads and an even number.<\/p>\n\n\n\n

P<\/mi>(<\/mo>heads<\/mtext>)<\/mo>=<\/mo>1<\/mn>2<\/mn><\/mfrac><\/mrow>P(\\text{heads})=\\frac{1}{2}<\/annotation><\/semantics><\/math><\/div>\n\n\n\n
P<\/mi>(<\/mo>even number<\/mtext>)<\/mo>=<\/mo>3<\/mn>6<\/mn><\/mfrac>=<\/mo>1<\/mn>2<\/mn><\/mfrac><\/mrow>P(\\text{even number})=\\frac{3}{6}=\\frac{1}{2}<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

The two experiments do not affect each other, so:<\/p>\n\n\n\n

P<\/mi>(<\/mo>heads and even<\/mtext>)<\/mo>=<\/mo>P<\/mi>(<\/mo>heads<\/mtext>)<\/mo>\u00d7<\/mo>P<\/mi>(<\/mo>even<\/mtext>)<\/mo><\/mrow>P(\\text{heads and even})=P(\\text{heads})\\times P(\\text{even})<\/annotation><\/semantics><\/math><\/div>\n\n\n\n
P<\/mi>(<\/mo>heads and even<\/mtext>)<\/mo>=<\/mo>1<\/mn>2<\/mn><\/mfrac>\u00d7<\/mo>1<\/mn>2<\/mn><\/mfrac><\/mrow>P(\\text{heads and even})=\\frac{1}{2}\\times\\frac{1}{2}<\/annotation><\/semantics><\/math><\/div>\n\n\n\n
P<\/mi>(<\/mo>heads and even<\/mtext>)<\/mo>=<\/mo>1<\/mn>4<\/mn><\/mfrac><\/mrow>P(\\text{heads and even})=\\frac{1}{4}<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

Events are dependent if the occurrence of the first event changes the probability of the second event. This usually happens when objects are selected without replacement.<\/p>\n\n\n\n

Example<\/strong><\/p>\n\n\n\n

A bag contains 3 red balls and 2 blue balls. Two balls are selected without replacement. Calculate the probability that both are red.<\/p>\n\n\n\n

For the first selection:<\/p>\n\n\n\n

P<\/mi>(<\/mo>R<\/mi>1<\/mn><\/msub>)<\/mo>=<\/mo>3<\/mn>5<\/mn><\/mfrac><\/mrow>P(R_1)=\\frac{3}{5}<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

After one red ball is selected, 2 red balls remain out of 4 balls:<\/p>\n\n\n\n

P<\/mi>(<\/mo>R<\/mi>2<\/mn><\/msub>|<\/mo>R<\/mi>1<\/mn><\/msub>)<\/mo>=<\/mo>2<\/mn>4<\/mn><\/mfrac><\/mrow>P(R_2\\mid R_1)=\\frac{2}{4}<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

Therefore:<\/p>\n\n\n\n

P<\/mi>(<\/mo>R<\/mi>1<\/mn><\/msub>\u2229<\/mo>R<\/mi>2<\/mn><\/msub>)<\/mo>=<\/mo>3<\/mn>5<\/mn><\/mfrac>\u00d7<\/mo>2<\/mn>4<\/mn><\/mfrac><\/mrow>P(R_1\\cap R_2)=\\frac{3}{5}\\times\\frac{2}{4}<\/annotation><\/semantics><\/math><\/div>\n\n\n\n
P<\/mi>(<\/mo>R<\/mi>1<\/mn><\/msub>\u2229<\/mo>R<\/mi>2<\/mn><\/msub>)<\/mo>=<\/mo>6<\/mn>20<\/mn><\/mfrac>=<\/mo>3<\/mn>10<\/mn><\/mfrac><\/mrow>P(R_1\\cap R_2)=\\frac{6}{20}=\\frac{3}{10}<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

Venn diagrams<\/h2>\n\n\n\n

A Venn diagram represents events as circles inside a rectangle representing the sample space. The overlap of two circles represents the intersection of the events.<\/p>\n\n\n\n

Important notation:<\/p>\n\n\n\n

A<\/mi>\u2229<\/mo>B<\/mi><\/mrow>A\\cap B<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

This means A and B.<\/p>\n\n\n\n

A<\/mi>\u222a<\/mo>B<\/mi><\/mrow>A\\cup B<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

This means A or B or both.<\/p>\n\n\n\n

A<\/mi>\u2032<\/mo><\/msup>A’<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

This means not A.<\/p>\n\n\n\n

Example<\/strong><\/p>\n\n\n\n

In a group of 40 learners, 22 study Mathematics, 18 study Physical Sciences and 10 study both subjects. Calculate the number who study at least one subject.<\/p>\n\n\n\n

Start by placing the intersection in the overlap.<\/p>\n\n\n\n

The number studying only Mathematics is:<\/p>\n\n\n\n

22<\/mn>\u2212<\/mo>10<\/mn>=<\/mo>12<\/mn><\/mrow>22-10=12<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

The number studying only Physical Sciences is:<\/p>\n\n\n\n

18<\/mn>\u2212<\/mo>10<\/mn>=<\/mo>8<\/mn><\/mrow>18-10=8<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

The number studying at least one subject is:<\/p>\n\n\n\n

n<\/mi>(<\/mo>M<\/mi>\u222a<\/mo>P<\/mi>)<\/mo>=<\/mo>12<\/mn>+<\/mo>10<\/mn>+<\/mo>8<\/mn><\/mrow>n(M\\cup P)=12+10+8<\/annotation><\/semantics><\/math><\/div>\n\n\n\n
n<\/mi>(<\/mo>M<\/mi>\u222a<\/mo>P<\/mi>)<\/mo>=<\/mo>30<\/mn><\/mrow>n(M\\cup P)=30<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

The number studying neither subject is:<\/p>\n\n\n\n

40<\/mn>\u2212<\/mo>30<\/mn>=<\/mo>10<\/mn><\/mrow>40-30=10<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

Remember<\/strong><\/p>\n\n\n\n

In a Venn diagram, fill in the intersection first. The totals for each circle include the intersection, so do not count the overlap twice.<\/p>\n\n\n\n

\"\"\/<\/a><\/figure>\n\n\n\n

Tree diagrams<\/h2>\n\n\n\n

A tree diagram shows the possible outcomes of consecutive events. Each branch is labelled with its probability.<\/p>\n\n\n\n

The probabilities leaving any point must add up to one. Multiply probabilities along a path and add the probabilities of different paths leading to the required event.<\/p>\n\n\n\n

Example<\/strong><\/p>\n\n\n\n

A bag contains 2 green balls and 3 yellow balls. Two balls are selected without replacement. Calculate the probability of selecting one ball of each colour.<\/p>\n\n\n\n

There are two possible paths: green then yellow, or yellow then green.<\/p>\n\n\n\n

First path:<\/p>\n\n\n\n

P<\/mi>(<\/mo>G<\/mi>Y<\/mi>)<\/mo>=<\/mo>2<\/mn>5<\/mn><\/mfrac>\u00d7<\/mo>3<\/mn>4<\/mn><\/mfrac><\/mrow>P(GY)=\\frac{2}{5}\\times\\frac{3}{4}<\/annotation><\/semantics><\/math><\/div>\n\n\n\n
P<\/mi>(<\/mo>G<\/mi>Y<\/mi>)<\/mo>=<\/mo>6<\/mn>20<\/mn><\/mfrac><\/mrow>P(GY)=\\frac{6}{20}<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

Second path:<\/p>\n\n\n\n

P<\/mi>(<\/mo>Y<\/mi>G<\/mi>)<\/mo>=<\/mo>3<\/mn>5<\/mn><\/mfrac>\u00d7<\/mo>2<\/mn>4<\/mn><\/mfrac><\/mrow>P(YG)=\\frac{3}{5}\\times\\frac{2}{4}<\/annotation><\/semantics><\/math><\/div>\n\n\n\n
P<\/mi>(<\/mo>Y<\/mi>G<\/mi>)<\/mo>=<\/mo>6<\/mn>20<\/mn><\/mfrac><\/mrow>P(YG)=\\frac{6}{20}<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

Add the two path probabilities:<\/p>\n\n\n\n

P<\/mi>(<\/mo>one of each<\/mtext>)<\/mo>=<\/mo>P<\/mi>(<\/mo>G<\/mi>Y<\/mi>)<\/mo>+<\/mo>P<\/mi>(<\/mo>Y<\/mi>G<\/mi>)<\/mo><\/mrow>P(\\text{one of each})=P(GY)+P(YG)<\/annotation><\/semantics><\/math><\/div>\n\n\n\n
P<\/mi>(<\/mo>one of each<\/mtext>)<\/mo>=<\/mo>6<\/mn>20<\/mn><\/mfrac>+<\/mo>6<\/mn>20<\/mn><\/mfrac><\/mrow>P(\\text{one of each})=\\frac{6}{20}+\\frac{6}{20}<\/annotation><\/semantics><\/math><\/div>\n\n\n\n
P<\/mi>(<\/mo>one of each<\/mtext>)<\/mo>=<\/mo>12<\/mn>20<\/mn><\/mfrac>=<\/mo>3<\/mn>5<\/mn><\/mfrac><\/mrow>P(\\text{one of each})=\\frac{12}{20}=\\frac{3}{5}<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

If selection is with replacement, the total number of objects and the branch probabilities remain unchanged.<\/p>\n\n\n\n

Contingency tables<\/h2>\n\n\n\n

A contingency table organises data according to two categories. The row and column totals are used to calculate probabilities.<\/p>\n\n\n\n

Example<\/strong><\/p>\n\n\n\n

A survey gives the following information:<\/p>\n\n\n\n

Of 50 learners, 30 are girls. Eighteen of the girls and 12 of the boys play sport.<\/p>\n\n\n\n

The number of boys is:<\/p>\n\n\n\n

50<\/mn>\u2212<\/mo>30<\/mn>=<\/mo>20<\/mn><\/mrow>50-30=20<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

The number of girls who do not play sport is:<\/p>\n\n\n\n

30<\/mn>\u2212<\/mo>18<\/mn>=<\/mo>12<\/mn><\/mrow>30-18=12<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

The number of boys who do not play sport is:<\/p>\n\n\n\n

20<\/mn>\u2212<\/mo>12<\/mn>=<\/mo>8<\/mn><\/mrow>20-12=8<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

The total number who play sport is:<\/p>\n\n\n\n

18<\/mn>+<\/mo>12<\/mn>=<\/mo>30<\/mn><\/mrow>18+12=30<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

The probability that a randomly selected learner is a girl who plays sport is:<\/p>\n\n\n\n

P<\/mi>(<\/mo>girl and sport<\/mtext>)<\/mo>=<\/mo>18<\/mn>50<\/mn><\/mfrac><\/mrow>P(\\text{girl and sport})=\\frac{18}{50}<\/annotation><\/semantics><\/math><\/div>\n\n\n\n
P<\/mi>(<\/mo>girl and sport<\/mtext>)<\/mo>=<\/mo>9<\/mn>25<\/mn><\/mfrac><\/mrow>P(\\text{girl and sport})=\\frac{9}{25}<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

The probability that a learner plays sport, given that the learner is a girl, is:<\/p>\n\n\n\n

P<\/mi>(<\/mo>sport<\/mtext>|<\/mo>girl<\/mtext>)<\/mo>=<\/mo>18<\/mn>30<\/mn><\/mfrac><\/mrow>P(\\text{sport}\\mid\\text{girl})=\\frac{18}{30}<\/annotation><\/semantics><\/math><\/div>\n\n\n\n
P<\/mi>(<\/mo>sport<\/mtext>|<\/mo>girl<\/mtext>)<\/mo>=<\/mo>3<\/mn>5<\/mn><\/mfrac><\/mrow>P(\\text{sport}\\mid\\text{girl})=\\frac{3}{5}<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

Exam Tip<\/strong><\/p>\n\n\n\n

For a probability involving \u201cgiven that\u201d, use the total of the stated group as the denominator.<\/p>\n\n\n\n

Addition rule<\/h2>\n\n\n\n

The general addition rule is used to calculate the probability that event A or event B occurs.<\/p>\n\n\n\n

P<\/mi>(<\/mo>A<\/mi>\u222a<\/mo>B<\/mi>)<\/mo>=<\/mo>P<\/mi>(<\/mo>A<\/mi>)<\/mo>+<\/mo>P<\/mi>(<\/mo>B<\/mi>)<\/mo>\u2212<\/mo>P<\/mi>(<\/mo>A<\/mi>\u2229<\/mo>B<\/mi>)<\/mo><\/mrow>P(A\\cup B)=P(A)+P(B)-P(A\\cap B)<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

The intersection is subtracted because it was counted once in each event and would otherwise be counted twice.<\/p>\n\n\n\n

Example<\/strong><\/p>\n\n\n\n

Suppose:<\/p>\n\n\n\n

P<\/mi>(<\/mo>A<\/mi>)<\/mo>=<\/mo>0,6<\/mn><\/mrow>P(A)=0{,}6<\/annotation><\/semantics><\/math><\/div>\n\n\n\n
P<\/mi>(<\/mo>B<\/mi>)<\/mo>=<\/mo>0,5<\/mn><\/mrow>P(B)=0{,}5<\/annotation><\/semantics><\/math><\/div>\n\n\n\n
P<\/mi>(<\/mo>A<\/mi>\u2229<\/mo>B<\/mi>)<\/mo>=<\/mo>0,2<\/mn><\/mrow>P(A\\cap B)=0{,}2<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

Then:<\/p>\n\n\n\n

P<\/mi>(<\/mo>A<\/mi>\u222a<\/mo>B<\/mi>)<\/mo>=<\/mo>0,6<\/mn>+<\/mo>0,5<\/mn>\u2212<\/mo>0,2<\/mn><\/mrow>P(A\\cup B)=0{,}6+0{,}5-0{,}2<\/annotation><\/semantics><\/math><\/div>\n\n\n\n
P<\/mi>(<\/mo>A<\/mi>\u222a<\/mo>B<\/mi>)<\/mo>=<\/mo>0,9<\/mn><\/mrow>P(A\\cup B)=0{,}9<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

If the events are mutually exclusive, the intersection is zero.<\/p>\n\n\n\n

Multiplication rule<\/h2>\n\n\n\n

The multiplication rule calculates the probability that events occur together.<\/p>\n\n\n\n

For independent events:<\/p>\n\n\n\n

P<\/mi>(<\/mo>A<\/mi>\u2229<\/mo>B<\/mi>)<\/mo>=<\/mo>P<\/mi>(<\/mo>A<\/mi>)<\/mo>\u00d7<\/mo>P<\/mi>(<\/mo>B<\/mi>)<\/mo><\/mrow>P(A\\cap B)=P(A)\\times P(B)<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

For dependent events:<\/p>\n\n\n\n

P<\/mi>(<\/mo>A<\/mi>\u2229<\/mo>B<\/mi>)<\/mo>=<\/mo>P<\/mi>(<\/mo>A<\/mi>)<\/mo>\u00d7<\/mo>P<\/mi>(<\/mo>B<\/mi>|<\/mo>A<\/mi>)<\/mo><\/mrow>P(A\\cap B)=P(A)\\times P(B\\mid A)<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

Example<\/strong><\/p>\n\n\n\n

The probability that a learner completes an assignment is 0,8. If the assignment is completed, the probability that it is submitted on time is 0,9. Calculate the probability that it is completed and submitted on time.<\/p>\n\n\n\n

P<\/mi>(<\/mo>C<\/mi>\u2229<\/mo>T<\/mi>)<\/mo>=<\/mo>P<\/mi>(<\/mo>C<\/mi>)<\/mo>\u00d7<\/mo>P<\/mi>(<\/mo>T<\/mi>|<\/mo>C<\/mi>)<\/mo><\/mrow>P(C\\cap T)=P(C)\\times P(T\\mid C)<\/annotation><\/semantics><\/math><\/div>\n\n\n\n
P<\/mi>(<\/mo>C<\/mi>\u2229<\/mo>T<\/mi>)<\/mo>=<\/mo>0,8<\/mn>\u00d7<\/mo>0,9<\/mn><\/mrow>P(C\\cap T)=0{,}8\\times 0{,}9<\/annotation><\/semantics><\/math><\/div>\n\n\n\n
P<\/mi>(<\/mo>C<\/mi>\u2229<\/mo>T<\/mi>)<\/mo>=<\/mo>0,72<\/mn><\/mrow>P(C\\cap T)=0{,}72<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

Remember<\/strong><\/p>\n\n\n\n

In probability language, \u201cand\u201d usually indicates multiplication or an intersection, while \u201cor\u201d usually indicates addition or a union.<\/p>\n\n\n\n

Fundamental counting principle<\/h2>\n\n\n\n

The fundamental counting principle determines the total number of possible arrangements or choices in a sequence of decisions.<\/p>\n\n\n\n

If the first decision has a certain number of choices and the second decision has another number of choices, multiply the numbers of choices.<\/p>\n\n\n\n

n<\/mi>(<\/mo>total outcomes<\/mtext>)<\/mo>=<\/mo>n<\/mi>1<\/mn><\/msub>\u00d7<\/mo>n<\/mi>2<\/mn><\/msub>\u00d7<\/mo>n<\/mi>3<\/mn><\/msub>\u00d7<\/mo>\u22ef<\/mo><\/mrow>n(\\text{total outcomes})=n_1\\times n_2\\times n_3\\times\\cdots<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

Example<\/strong><\/p>\n\n\n\n

A meal consists of one main course, one drink and one dessert. There are 4 main courses, 3 drinks and 2 desserts.<\/p>\n\n\n\n

n<\/mi>(<\/mo>meals<\/mtext>)<\/mo>=<\/mo>4<\/mn>\u00d7<\/mo>3<\/mn>\u00d7<\/mo>2<\/mn><\/mrow>n(\\text{meals})=4\\times 3\\times 2<\/annotation><\/semantics><\/math><\/div>\n\n\n\n
n<\/mi>(<\/mo>meals<\/mtext>)<\/mo>=<\/mo>24<\/mn><\/mrow>n(\\text{meals})=24<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

Example<\/strong><\/p>\n\n\n\n

A four-digit code is formed from the digits 1 to 6 without repetition.<\/p>\n\n\n\n

There are 6 choices for the first position, followed by 5, 4 and 3 choices.<\/p>\n\n\n\n

n<\/mi>(<\/mo>codes<\/mtext>)<\/mo>=<\/mo>6<\/mn>\u00d7<\/mo>5<\/mn>\u00d7<\/mo>4<\/mn>\u00d7<\/mo>3<\/mn><\/mrow>n(\\text{codes})=6\\times 5\\times 4\\times 3<\/annotation><\/semantics><\/math><\/div>\n\n\n\n
n<\/mi>(<\/mo>codes<\/mtext>)<\/mo>=<\/mo>360<\/mn><\/mrow>n(\\text{codes})=360<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

If repetition is allowed:<\/p>\n\n\n\n

n<\/mi>(<\/mo>codes<\/mtext>)<\/mo>=<\/mo>6<\/mn>\u00d7<\/mo>6<\/mn>\u00d7<\/mo>6<\/mn>\u00d7<\/mo>6<\/mn><\/mrow>n(\\text{codes})=6\\times 6\\times 6\\times 6<\/annotation><\/semantics><\/math><\/div>\n\n\n\n
n<\/mi>(<\/mo>codes<\/mtext>)<\/mo>=<\/mo>1296<\/mn><\/mrow>n(\\text{codes})=1296<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

Common Mistake<\/strong><\/p>\n\n\n\n

Apply restrictions before multiplying. For example, the first digit of a number cannot be zero. If a five-digit number is formed from the digits 0 to 7 without repetition, there are only 7 choices for the first digit, not 8.<\/p>\n\n\n\n

n<\/mi>(<\/mo>five-digit numbers<\/mtext>)<\/mo>=<\/mo>7<\/mn>\u00d7<\/mo>7<\/mn>\u00d7<\/mo>6<\/mn>\u00d7<\/mo>5<\/mn>\u00d7<\/mo>4<\/mn><\/mrow>n(\\text{five-digit numbers})=7\\times 7\\times 6\\times 5\\times 4<\/annotation><\/semantics><\/math><\/div>\n\n\n\n
n<\/mi>(<\/mo>five-digit numbers<\/mtext>)<\/mo>=<\/mo>5880<\/mn><\/mrow>n(\\text{five-digit numbers})=5880<\/annotation><\/semantics><\/math><\/div>\n\n\n\n

Final Exam Tip<\/strong><\/p>\n\n\n\n

Identify key words before choosing a rule. Use the complement for \u201cnot\u201d or \u201cat least one\u201d, the addition rule for \u201cor\u201d, the multiplication rule for \u201cand\u201d, and the fundamental counting principle when counting arrangements. Always check that the final probability is between zero and one.<\/p>\n","protected":false},"excerpt":{"rendered":"

Probability Probability measures how likely an event is to occur. It is used to analyse uncertainty and make predictions. In examinations, show the correct formula, substitute values clearly, simplify accurately and give the final answer. This method-based approach is important because marks are awarded for correct mathematical steps as well as the answer. Basic probability […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_acf_changed":false,"_breakdance_hide_in_design_set":false,"_breakdance_tags":"","footnotes":""},"categories":[1],"tags":[],"class_list":["post-104","post","type-post","status-publish","format-standard","hentry","category-uncategorized"],"acf":{"document_name":"Probability","grade":12,"subject":"Mathematics","term":4,"paper":1},"_links":{"self":[{"href":"https:\/\/repstaq.com\/study-guide-docs\/wp-json\/wp\/v2\/posts\/104","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/repstaq.com\/study-guide-docs\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/repstaq.com\/study-guide-docs\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/repstaq.com\/study-guide-docs\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/repstaq.com\/study-guide-docs\/wp-json\/wp\/v2\/comments?post=104"}],"version-history":[{"count":4,"href":"https:\/\/repstaq.com\/study-guide-docs\/wp-json\/wp\/v2\/posts\/104\/revisions"}],"predecessor-version":[{"id":241,"href":"https:\/\/repstaq.com\/study-guide-docs\/wp-json\/wp\/v2\/posts\/104\/revisions\/241"}],"wp:attachment":[{"href":"https:\/\/repstaq.com\/study-guide-docs\/wp-json\/wp\/v2\/media?parent=104"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/repstaq.com\/study-guide-docs\/wp-json\/wp\/v2\/categories?post=104"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/repstaq.com\/study-guide-docs\/wp-json\/wp\/v2\/tags?post=104"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}