{"id":259,"date":"2026-09-09T06:39:17","date_gmt":"2026-09-09T06:39:17","guid":{"rendered":"https:\/\/repstaq.com\/study-guide-docs\/probability-g11-maths-t4p1\/"},"modified":"2026-09-09T06:39:17","modified_gmt":"2026-09-09T06:39:17","slug":"probability-g11-maths-t4p1","status":"publish","type":"post","link":"https:\/\/repstaq.com\/study-guide-docs\/probability-g11-maths-t4p1\/","title":{"rendered":"Probability-G11-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 to make predictions in situations involving games, surveys, weather, insurance, medicine and quality control.<\/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

For equally likely outcomes:<\/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><\/math><\/div>\n\n\n\n

Here, the number of outcomes in event A is compared with the total number of outcomes in the sample space.<\/p>\n\n\n\n

A probability is always between 0 and 1:<\/p>\n\n\n\n

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

An impossible event has probability 0, while a certain event has probability 1.<\/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 multiple of 3.<\/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><\/math><\/div>\n\n\n\n

The multiples of 3 are:<\/p>\n\n\n\n

A<\/mi>=<\/mo>{<\/mo>3<\/mn>;<\/mo>6<\/mn>}<\/mo><\/mrow><\/math><\/div>\n\n\n\n

Therefore:<\/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><\/math><\/div>\n\n\n\n
P<\/mi>(<\/mo>A<\/mi>)<\/mo>=<\/mo>2<\/mn>6<\/mn><\/mfrac><\/mrow><\/math><\/div>\n\n\n\n
P<\/mi>(<\/mo>A<\/mi>)<\/mo>=<\/mo>1<\/mn>3<\/mn><\/mfrac><\/mrow><\/math><\/div>\n\n\n\n

Relative frequency versus theoretical probability<\/h2>\n\n\n\n

Theoretical probability is calculated from all possible equally likely outcomes. Relative frequency is based on results obtained from an experiment or collected data.<\/p>\n\n\n\n

Relative frequency<\/mtext>=<\/mo>Number of times the event occurs<\/mtext>Total number of trials<\/mtext><\/mfrac><\/mrow><\/math><\/div>\n\n\n\n

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

A coin is tossed 80 times and lands on heads 46 times. Calculate the relative frequency of heads.<\/p>\n\n\n\n

Relative frequency of heads<\/mtext>=<\/mo>46<\/mn>80<\/mn><\/mfrac><\/mrow><\/math><\/div>\n\n\n\n
Relative frequency of heads<\/mtext>=<\/mo>0,575<\/mn><\/mrow><\/math><\/div>\n\n\n\n

For a fair coin, the theoretical probability of heads is:<\/p>\n\n\n\n

P<\/mi>(<\/mo>heads<\/mtext>)<\/mo>=<\/mo>1<\/mn>2<\/mn><\/mfrac>=<\/mo>0,5<\/mn><\/mrow><\/math><\/div>\n\n\n\n

Experimental results do not always equal theoretical probability. However, as the number of trials increases, the relative frequency usually approaches the theoretical probability.<\/p>\n\n\n\n

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

The complement of event A, written as A prime, 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><\/math><\/div>\n\n\n\n

Also:<\/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><\/math><\/div>\n\n\n\n

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

The probability that a learner passes a test is 0,78. Calculate the probability that the learner does not pass.<\/p>\n\n\n\n

P<\/mi>(<\/mo>not passing<\/mtext>)<\/mo>=<\/mo>1<\/mn>\u2212<\/mo>P<\/mi>(<\/mo>passing<\/mtext>)<\/mo><\/mrow><\/math><\/div>\n\n\n\n
P<\/mi>(<\/mo>not passing<\/mtext>)<\/mo>=<\/mo>1<\/mn>\u2212<\/mo>0,78<\/mn><\/mrow><\/math><\/div>\n\n\n\n
P<\/mi>(<\/mo>not passing<\/mtext>)<\/mo>=<\/mo>0,22<\/mn><\/mrow><\/math><\/div>\n\n\n\n

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

\u201cAt least one\u201d questions are often easier to solve by first calculating the probability of none occurring.<\/p>\n\n\n\n

P<\/mi>(<\/mo>at least one<\/mtext>)<\/mo>=<\/mo>1<\/mn>\u2212<\/mo>P<\/mi>(<\/mo>none<\/mtext>)<\/mo><\/mrow><\/math><\/div>\n\n\n\n

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

Mutually exclusive events 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><\/math><\/div>\n\n\n\n

For example, when one die is rolled, obtaining an even number and obtaining an odd number are mutually exclusive events.<\/p>\n\n\n\n

If events are mutually exclusive:<\/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><\/math><\/div>\n\n\n\n

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

Mutually exclusive does not mean independent. If one of two mutually exclusive events occurs, the other event cannot occur.<\/p>\n\n\n\n

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

Two events are independent if the occurrence of one event does not affect the probability of the other event.<\/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><\/math><\/div>\n\n\n\n

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

P<\/mi>(<\/mo>A<\/mi>|<\/mo>B<\/mi>)<\/mo>=<\/mo>P<\/mi>(<\/mo>A<\/mi>)<\/mo><\/mrow><\/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 tails and a number greater than 4.<\/p>\n\n\n\n

P<\/mi>(<\/mo>tails<\/mtext>)<\/mo>=<\/mo>1<\/mn>2<\/mn><\/mfrac><\/mrow><\/math><\/div>\n\n\n\n
P<\/mi>(<\/mo>number greater than <\/mtext>4<\/mn>)<\/mo>=<\/mo>2<\/mn>6<\/mn><\/mfrac>=<\/mo>1<\/mn>3<\/mn><\/mfrac><\/mrow><\/math><\/div>\n\n\n\n

The events are independent, so:<\/p>\n\n\n\n

P<\/mi>(<\/mo>tails and number greater than <\/mtext>4<\/mn>)<\/mo>=<\/mo>1<\/mn>2<\/mn><\/mfrac>\u00d7<\/mo>1<\/mn>3<\/mn><\/mfrac><\/mrow><\/math><\/div>\n\n\n\n
P<\/mi>(<\/mo>tails and number greater than <\/mtext>4<\/mn>)<\/mo>=<\/mo>1<\/mn>6<\/mn><\/mfrac><\/mrow><\/math><\/div>\n\n\n\n

Dependent events<\/h2>\n\n\n\n

Events are dependent when 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

Conditional probability is written as:<\/p>\n\n\n\n

P<\/mi>(<\/mo>B<\/mi>|<\/mo>A<\/mi>)<\/mo>=<\/mo>P<\/mi>(<\/mo>A<\/mi>\u2229<\/mo>B<\/mi>)<\/mo><\/mrow>P<\/mi>(<\/mo>A<\/mi>)<\/mo><\/mrow><\/mfrac><\/mrow><\/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><\/math><\/div>\n\n\n\n

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

A bag contains 5 red balls and 3 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>first red<\/mtext>)<\/mo>=<\/mo>5<\/mn>8<\/mn><\/mfrac><\/mrow><\/math><\/div>\n\n\n\n

After a red ball has been selected, 4 red balls remain out of 7 balls.<\/p>\n\n\n\n

P<\/mi>(<\/mo>second red<\/mtext>|<\/mo>first red<\/mtext>)<\/mo>=<\/mo>4<\/mn>7<\/mn><\/mfrac><\/mrow><\/math><\/div>\n\n\n\n

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

P<\/mi>(<\/mo>both red<\/mtext>)<\/mo>=<\/mo>5<\/mn>8<\/mn><\/mfrac>\u00d7<\/mo>4<\/mn>7<\/mn><\/mfrac><\/mrow><\/math><\/div>\n\n\n\n
P<\/mi>(<\/mo>both red<\/mtext>)<\/mo>=<\/mo>20<\/mn>56<\/mn><\/mfrac><\/mrow><\/math><\/div>\n\n\n\n
P<\/mi>(<\/mo>both red<\/mtext>)<\/mo>=<\/mo>5<\/mn>14<\/mn><\/mfrac><\/mrow><\/math><\/div>\n\n\n\n\n
\n\n\"\"\n<\/a>\n<\/figure>\n\n\n\n

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

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

For any two events:<\/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><\/math><\/div>\n\n\n\n

The intersection is subtracted because it was included twice.<\/p>\n\n\n\n

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

In a class, 18 learners take Mathematics, 15 take Physical Sciences, and 10 take both subjects. There are 30 learners. Calculate the probability that a randomly selected learner takes Mathematics or Physical Sciences.<\/p>\n\n\n\n

P<\/mi>(<\/mo>M<\/mi>\u222a<\/mo>P<\/mi>)<\/mo>=<\/mo>P<\/mi>(<\/mo>M<\/mi>)<\/mo>+<\/mo>P<\/mi>(<\/mo>P<\/mi>)<\/mo>\u2212<\/mo>P<\/mi>(<\/mo>M<\/mi>\u2229<\/mo>P<\/mi>)<\/mo><\/mrow><\/math><\/div>\n\n\n\n
P<\/mi>(<\/mo>M<\/mi>\u222a<\/mo>P<\/mi>)<\/mo>=<\/mo>18<\/mn>30<\/mn><\/mfrac>+<\/mo>15<\/mn>30<\/mn><\/mfrac>\u2212<\/mo>10<\/mn>30<\/mn><\/mfrac><\/mrow><\/math><\/div>\n\n\n\n
P<\/mi>(<\/mo>M<\/mi>\u222a<\/mo>P<\/mi>)<\/mo>=<\/mo>23<\/mn>30<\/mn><\/mfrac><\/mrow><\/math><\/div>\n\n\n\n

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

The multiplication rule is used when events are joined by \u201cand\u201d.<\/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><\/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><\/math><\/div>\n\n\n\n

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

\u201cAnd\u201d usually indicates multiplication or an intersection. \u201cOr\u201d usually indicates addition or a union. Always consider whether the events overlap, are independent, or are dependent before selecting a formula.<\/p>\n\n\n\n

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

A two-event Venn diagram consists of two circles inside a rectangle. The rectangle represents the universal set. The overlapping region represents the intersection of the two events.<\/p>\n\n\n\n

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

All elements in A, in B, or in both are represented by:<\/p>\n\n\n\n

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

Elements outside both circles are represented by:<\/p>\n\n\n\n

(<\/mo>A<\/mi>\u222a<\/mo>B<\/mi>)<\/mo><\/mrow>\u2032<\/mo><\/msup><\/math><\/div>\n\n\n\n

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

In a group of 40 learners, 22 play soccer, 17 play netball and 8 play both. Determine how many play neither sport.<\/p>\n\n\n\n

Place the intersection first:<\/p>\n\n\n\n

n<\/mi>(<\/mo>S<\/mi>\u2229<\/mo>N<\/mi>)<\/mo>=<\/mo>8<\/mn><\/mrow><\/math><\/div>\n\n\n\n

The number who play only soccer is:<\/p>\n\n\n\n

22<\/mn>\u2212<\/mo>8<\/mn>=<\/mo>14<\/mn><\/mrow><\/math><\/div>\n\n\n\n

The number who play only netball is:<\/p>\n\n\n\n

17<\/mn>\u2212<\/mo>8<\/mn>=<\/mo>9<\/mn><\/mrow><\/math><\/div>\n\n\n\n

The number who play at least one sport is:<\/p>\n\n\n\n

n<\/mi>(<\/mo>S<\/mi>\u222a<\/mo>N<\/mi>)<\/mo>=<\/mo>14<\/mn>+<\/mo>8<\/mn>+<\/mo>9<\/mn><\/mrow><\/math><\/div>\n\n\n\n
n<\/mi>(<\/mo>S<\/mi>\u222a<\/mo>N<\/mi>)<\/mo>=<\/mo>31<\/mn><\/mrow><\/math><\/div>\n\n\n\n

Therefore, the number who play neither sport is:<\/p>\n\n\n\n

40<\/mn>\u2212<\/mo>31<\/mn>=<\/mo>9<\/mn><\/mrow><\/math><\/div>\n\n\n\n

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

A three-event Venn diagram has three overlapping circles. Begin by filling in the intersection common to all three events.<\/p>\n\n\n\n

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

P<\/mi>(<\/mo>A<\/mi>\u222a<\/mo>B<\/mi>\u222a<\/mo>C<\/mi>)<\/mo>=<\/mo>P<\/mi>(<\/mo>A<\/mi>)<\/mo>+<\/mo>P<\/mi>(<\/mo>B<\/mi>)<\/mo>+<\/mo>P<\/mi>(<\/mo>C<\/mi>)<\/mo>\u2212<\/mo>P<\/mi>(<\/mo>A<\/mi>\u2229<\/mo>B<\/mi>)<\/mo>\u2212<\/mo>P<\/mi>(<\/mo>A<\/mi>\u2229<\/mo>C<\/mi>)<\/mo>\u2212<\/mo>P<\/mi>(<\/mo>B<\/mi>\u2229<\/mo>C<\/mi>)<\/mo>+<\/mo>P<\/mi>(<\/mo>A<\/mi>\u2229<\/mo>B<\/mi>\u2229<\/mo>C<\/mi>)<\/mo><\/mrow><\/math><\/div>\n\n\n\n

When completing a three-event diagram, use this order:<\/p>\n\n\n\n

First place the number in the intersection of all three events. Then calculate the parts belonging to exactly two events. Next calculate the parts belonging to only one event. Finally calculate the number outside all three circles.<\/p>\n\n\n\n

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

Suppose 6 learners take all three subjects, while 12 take Mathematics and Physical Sciences, including those who take all three. The number taking exactly Mathematics and Physical Sciences is:<\/p>\n\n\n\n

12<\/mn>\u2212<\/mo>6<\/mn>=<\/mo>6<\/mn><\/mrow><\/math><\/div>\n\n\n\n

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

Unless the question says \u201conly\u201d or \u201cexactly\u201d, the number in a two-event intersection may include the learners in the intersection of all three events.<\/p>\n\n\n\n

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

A tree diagram displays the possible outcomes of two or more stages. Probabilities are written on branches.<\/p>\n\n\n\n

At each stage, the branch probabilities must add to 1:<\/p>\n\n\n\n

\u2211<\/mo>P<\/mi>(<\/mo>branches<\/mtext>)<\/mo>=<\/mo>1<\/mn><\/mrow><\/math><\/div>\n\n\n\n

Multiply probabilities along a path and add the probabilities of different paths that produce the required result.<\/p>\n\n\n\n

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

A bag contains 2 green counters and 3 yellow counters. Two counters are selected with replacement. Calculate the probability of selecting one counter of each colour.<\/p>\n\n\n\n

One possible order is green then yellow:<\/p>\n\n\n\n

P<\/mi>(<\/mo>G<\/mi>Y<\/mi>)<\/mo>=<\/mo>2<\/mn>5<\/mn><\/mfrac>\u00d7<\/mo>3<\/mn>5<\/mn><\/mfrac><\/mrow><\/math><\/div>\n\n\n\n
P<\/mi>(<\/mo>G<\/mi>Y<\/mi>)<\/mo>=<\/mo>6<\/mn>25<\/mn><\/mfrac><\/mrow><\/math><\/div>\n\n\n\n

The other possible order is yellow then green:<\/p>\n\n\n\n

P<\/mi>(<\/mo>Y<\/mi>G<\/mi>)<\/mo>=<\/mo>3<\/mn>5<\/mn><\/mfrac>\u00d7<\/mo>2<\/mn>5<\/mn><\/mfrac><\/mrow><\/math><\/div>\n\n\n\n
P<\/mi>(<\/mo>Y<\/mi>G<\/mi>)<\/mo>=<\/mo>6<\/mn>25<\/mn><\/mfrac><\/mrow><\/math><\/div>\n\n\n\n

Add the two possible paths:<\/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><\/math><\/div>\n\n\n\n
P<\/mi>(<\/mo>one of each<\/mtext>)<\/mo>=<\/mo>6<\/mn>25<\/mn><\/mfrac>+<\/mo>6<\/mn>25<\/mn><\/mfrac><\/mrow><\/math><\/div>\n\n\n\n
P<\/mi>(<\/mo>one of each<\/mtext>)<\/mo>=<\/mo>12<\/mn>25<\/mn><\/mfrac><\/mrow><\/math><\/div>\n\n\n\n

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

A contingency table classifies data according to two categories. It includes row totals, column totals and a grand total.<\/p>\n\n\n\n

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

A survey includes 50 learners. Of the 28 girls, 18 prefer tea. Of the 22 boys, 12 prefer tea.<\/p>\n\n\n\n

The total number who prefer tea is:<\/p>\n\n\n\n

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

The number who do not prefer tea is:<\/p>\n\n\n\n

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

The probability that a randomly selected learner prefers tea is:<\/p>\n\n\n\n

P<\/mi>(<\/mo>tea<\/mtext>)<\/mo>=<\/mo>30<\/mn>50<\/mn><\/mfrac>=<\/mo>3<\/mn>5<\/mn><\/mfrac><\/mrow><\/math><\/div>\n\n\n\n

The probability that the learner is a girl and prefers tea is:<\/p>\n\n\n\n

P<\/mi>(<\/mo>girl and tea<\/mtext>)<\/mo>=<\/mo>18<\/mn>50<\/mn><\/mfrac>=<\/mo>9<\/mn>25<\/mn><\/mfrac><\/mrow><\/math><\/div>\n\n\n\n

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

P<\/mi>(<\/mo>tea<\/mtext>|<\/mo>girl<\/mtext>)<\/mo>=<\/mo>18<\/mn>28<\/mn><\/mfrac>=<\/mo>9<\/mn>14<\/mn><\/mfrac><\/mrow><\/math><\/div>\n\n\n\n

For conditional probability, the denominator must be the total of the group stated after the word \u201cgiven\u201d.<\/p>\n\n\n\n

Real life probability problems<\/h2>\n\n\n\n

Probability is used in weather forecasting, medical testing, insurance, manufacturing, transport and opinion surveys. Real-life questions often require learners to identify whether events are complementary, mutually exclusive, independent or dependent.<\/p>\n\n\n\n

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

A factory reports that 4% of its light bulbs are defective. Two bulbs are selected independently. Calculate the probability that at least one is defective.<\/p>\n\n\n\n

First calculate the probability that a bulb is not defective:<\/p>\n\n\n\n

P<\/mi>(<\/mo>not defective<\/mtext>)<\/mo>=<\/mo>1<\/mn>\u2212<\/mo>0,04<\/mn><\/mrow><\/math><\/div>\n\n\n\n
P<\/mi>(<\/mo>not defective<\/mtext>)<\/mo>=<\/mo>0,96<\/mn><\/mrow><\/math><\/div>\n\n\n\n

Calculate the probability that neither bulb is defective:<\/p>\n\n\n\n

P<\/mi>(<\/mo>neither defective<\/mtext>)<\/mo>=<\/mo>0,96<\/mn>\u00d7<\/mo>0,96<\/mn><\/mrow><\/math><\/div>\n\n\n\n
P<\/mi>(<\/mo>neither defective<\/mtext>)<\/mo>=<\/mo>0,9216<\/mn><\/mrow><\/math><\/div>\n\n\n\n

Use the complementary event:<\/p>\n\n\n\n

P<\/mi>(<\/mo>at least one defective<\/mtext>)<\/mo>=<\/mo>1<\/mn>\u2212<\/mo>0,9216<\/mn><\/mrow><\/math><\/div>\n\n\n\n
P<\/mi>(<\/mo>at least one defective<\/mtext>)<\/mo>=<\/mo>0,0784<\/mn><\/mrow><\/math><\/div>\n\n\n\n
P<\/mi>(<\/mo>at least one defective<\/mtext>)<\/mo>=<\/mo>7,84<\/mn>%<\/mo><\/mrow><\/math><\/div>\n\n\n\n

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

Read the wording carefully. \u201cWithout replacement\u201d normally creates dependent events, while \u201cwith replacement\u201d normally creates independent events. Show the formula, substitute correctly, simplify the answer, and check that the final probability lies between 0 and 1.<\/p>\n\n\n\n\n

\n\n\"\"\n<\/a>\n<\/figure>\n","protected":false},"excerpt":{"rendered":"

Probability Probability measures how likely an event is to occur. It is used to analyse uncertainty and to make predictions in situations involving games, surveys, weather, insurance, medicine and quality control. Basic probability An experiment is a process with an uncertain result, such as tossing a coin. An outcome is one possible result, while the […]<\/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-259","post","type-post","status-publish","format-standard","hentry","category-uncategorized"],"acf":{"document_name":"Probability","grade":11,"subject":"Mathematics","term":4,"paper":1},"_links":{"self":[{"href":"https:\/\/repstaq.com\/study-guide-docs\/wp-json\/wp\/v2\/posts\/259","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=259"}],"version-history":[{"count":0,"href":"https:\/\/repstaq.com\/study-guide-docs\/wp-json\/wp\/v2\/posts\/259\/revisions"}],"wp:attachment":[{"href":"https:\/\/repstaq.com\/study-guide-docs\/wp-json\/wp\/v2\/media?parent=259"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/repstaq.com\/study-guide-docs\/wp-json\/wp\/v2\/categories?post=259"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/repstaq.com\/study-guide-docs\/wp-json\/wp\/v2\/tags?post=259"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}