Wins out of all the chances
A spinner has 4 equal sections, labeled a, b, c and d. The sections are the same size, so the pointer is equally likely to stop on each one. Say that a wins and the other three sections lose.
Four equal sections, and only a wins.
The fraction that wins
Count the sections that win, and count all the sections: 1 section wins out of 4. The probability of winning is that fraction, .
This works whenever every outcome is equally likely. An outcome that counts as a win is called a favorable outcome, and the probability of the event is the number of favorable outcomes ÷ the total number of outcomes.
1 of the 4 equal parts wins, so the probability of winning is .
More than one favorable outcome
A fair dice has 6 equally likely outcomes. Three of them, 2, 4 and 6, are even numbers, so the probability of rolling an even number is , which simplifies to .
The probability of rolling a number greater than 4 is , because only two outcomes, 5 and 6, are greater than 4.
The 6 outcomes of a dice, with the 3 even ones colored: 3 out of 6.
Only equal sections can be counted
Counting works only when every outcome is equally likely. This spinner also has four sections, a, b, c and d, but d covers half of the disc. The pointer stops on d far more often than on a, so the probability of a is not .
The four sections are not the same size: d covers half of the disc.
To count, first cut the disc into equal sections. a, b and c are each one sixth of the disc. d is half of the disc, which is three sixths. Now the disc is 6 equal sections, and a is 1 of them, so the probability of a is . d is 3 of them, so the probability of d is .
The same disc cut into 6 equal sections. d fills 3 of them and a fills 1, so the probability of a is .
All the outcomes go on the bottom
Two mistakes are common. One is to put only the losing outcomes on the bottom. With 1 winning section and 3 losing ones, the answer is not : the bottom number counts every outcome, winning and losing together, and that is 4.
The other mistake is to turn the fraction over. is 4, which is more than 1, and no probability is more than 1. The favorable outcomes go on top.
Worked example: A Prize Wheel at a Carnival, Where One Section Fills Half the Wheel
Question A prize wheel at a carnival has five sections. The yellow section is half of the wheel. The other half is cut into four equal sections: two red, one blue and one green. The pointer is equally likely to stop anywhere on the wheel. (a) What is the probability that the pointer stops on red? (b) A player wins a prize when the pointer stops on yellow or on green. What is the probability of winning a prize?
1.Each small section is one quarter of a half, so it is 18 of the wheel. The yellow half is the same size as 4 of these small sections.
Each small section is one quarter of a half: 18 of the wheel. 2.Now the wheel is 8 equal sections: 4 yellow, 2 red, 1 blue and 1 green. Each of the 8 is equally likely.
Cut the yellow half into 4 small sections too. Now there are 8 equal sections. 3.(a) Red is 2 of the 8 equal sections, so the probability of red is 28 = 14.
(a) Red is 2 of the 8 equal sections: 28 = 14. 4.(b) Yellow or green is 4 + 1 = 5 of the 8 equal sections, so the probability of winning a prize is 58.
(b) Yellow or green is 5 of the 8 equal sections: 58.
Answer: (a) 14; (b) 58
Common mistakes
- Saying the probability of red is 25, because 2 of the 5 sections are red. The sections are different sizes, so they are not equally likely: the yellow section is as big as four small ones.
- Giving yellow a probability of 15 in part (b). Yellow is half the wheel, 48, so yellow or green is 48 + 18 = 58.
Worked example: A Bag of Marbles at the School Fair, and Green Marbles Added for an Even Chance
Question A bag at the school fair holds 4 red, 5 blue and 3 green marbles of the same size. Nina takes one marble without looking. (a) What is the probability that her marble is blue? (b) Nina puts her marble back. The stall keeper then adds some green marbles, and no other marbles, so that the probability of taking a green marble becomes 12. How many green marbles does he add?
1.Count all the marbles: 4 + 5 + 3 = 12. Each of the 12 marbles is equally likely to be taken.
There are 4 + 5 + 3 = 12 marbles, and each is equally likely to be taken. 2.(a) 5 of the 12 marbles are blue, so the probability of blue is 512.
(a) 5 of the 12 marbles are blue: 512. 3.A probability of 12 means that half the marbles are green. Then there must be as many green marbles as all the other marbles together.
For a probability of 12, the green marbles must match all the others. 4.The other marbles are 4 red and 5 blue, which is 9. Adding green marbles does not change them, so the bag needs 9 green marbles.
The others are 4 + 5 = 9, so the bag needs 9 green marbles. 5.(b) There are 3 green marbles already, so he adds 9 − 3 = 6. Check: the bag now has 18 marbles, and 918 = 12.
(b) He adds 9 − 3 = 6 green marbles, and 918 = 12.
Answer: (a) 512; (b) 6 green marbles
Common mistakes
- Writing the probability of blue as 57, the blue marbles against the marbles that are not blue. A probability compares the marbles that win with all 12 marbles in the bag.
- Adding 3 green marbles to make 6, because 6 is half of 12. The new marbles are in the bag too, so the total grows to 15, and 615 is not 12.