Probability · Grade 10 · 4 units
In Probability (Grade 10 · 4 units) you practise 8 skills: Events · The rules of probability · Venn diagrams · Conditional probability · Probability tables · Probability trees · Total probability and Bayes’ rule · Independent events. The probability of an event is a number from zero to one. The probability of the complement is one minus the probability of the event. When events are mutually exclusive, so they cannot happen together, the probability that one of them happens is the sum of their probabilities. When they are not, subtract the probability that both happen, because it was counted twice.
A conditional probability is the probability of an event when you know another event happened: divide the probability that both happen by the probability of the known event. In a table you see it by looking only at the known event’s row or column, and in a tree it is the probability on the second branch. When you know the conditional probabilities and want the reverse one, first work out the total probability, then divide. Two events are independent when knowing that one happened does not change the probability of the other. Check it in one of three equivalent ways: the probability of both together equals the product of the probabilities, or a conditional probability equals the plain one, in either direction.
Tip: Write in words what is known and what is asked before you choose a formula: the word “given” says it is a conditional probability.
What you learn
Events
For example: A dart lands at random on the board, with the same chance of hitting any point. What is the probability that it hits the shaded region?
Answer and solution
The answer: 1/10
The shaded cells: 2 × 5 = 10 . On the whole board: 10 × 10 = 100 . The probability: 10/100 = 1/10
The rules of probability
For example: Each pupil in a school year chose one favourite colour. A pupil is picked at random. The probability that the pupil chose red is 0.4 and blue 0.1 . What is the probability that the pupil chose red or blue?
Answer and solution
The answer: 0.5
Each pupil chose only one colour, so the events are mutually exclusive, and you add: 0.4 + 0.1 = 0.5
Venn diagrams
For example: A class has 25 pupils. The diagram shows how many of them play football, how many basketball, how many both and how many neither.
A pupil is picked at random. What is the probability that the pupil plays both football and basketball? Give the answer as a fraction.Answer and solution
The answer: 12/25
The overlap of the two circles holds 12 pupils, so the probability is 12/25
Conditional probability
For example: The table shows the pupils in a school year by how they came to school and whether they arrived on time.
| On time | Late | Total | |
|---|---|---|---|
| By bus | 5 | 20 | 25 |
| Not by bus | 5 | 11 | 16 |
| Total | 10 | 31 | 41 |
Answer and solution
The answer: 1/2
Look only at the column of the pupils who arrived on time: 5 + 5 = 10 . Of them, by bus: 5 , so the probability is 5/10 = 1/2
Probability tables
For example:
| First / second | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
| 1 | 2 | 3 | 4 | 5 |
| 2 | 3 | 4 | 5 | 6 |
| 3 | 4 | 5 | 6 | 7 |
| 4 | 5 | 6 | 7 | 8 |
Answer and solution
The answer: 1/8
The number of all the outcomes: 4 × 4 = 16 . The ones you want: (1, 2) , (2, 1) : 2 outcomes. The probability: 2/16 = 1/8
Probability trees
For example: A bag has 6 red balls and 2 blue balls. A ball is taken out, its colour noted, and it is put back. Then a second ball is taken out. What is the probability that both balls are red?
Answer and solution
The answer: 9/16
First stage: of the 8 balls, 6 are red, so the probability of a red ball is 6/8 . In the second stage the ball is back, so the probability is again 6/8 . Multiply along the branch: 6/8 × 6/8 = 36/64 = 9/16
Total probability and Bayes’ rule
For example: In a factory two machines make all the items. An item is picked at random.
| Machine | Share of the items | Probability of a faulty item |
|---|---|---|
| Machine A | 0.2 | 0.09 |
| Machine B | 0.8 | 0.02 |
Answer and solution
The answer: 0.034
Add the two branches: 0.2 × 0.09 + 0.8 × 0.02 = 0.018 + 0.016 = 0.034
Independent events
For example: The event A and the event B are independent. Given: P(A) = 0.2 and P(B) = 0.9 . Find P(A ∩ B) .
Answer and solution
The answer: 0.18
P(A ∩ B) = 0.2 × 0.9 = 0.18
Not on the site yet
Parts of this topic that are not built yet: Proving the three definitions are equivalent · Running experiments · Geometric probability.
The exercises are made by rules and the answers are worked out in code, so there are new exercises every time. The topic follows the Israeli Ministry of Education’s high school maths curriculum at 4 units (grades 10 to 12).