FP2–FP4, HP2–HP4 - Probability and expected outcomes

FP2–FP4, HP2–HP4 - Probability and expected outcomes

Describe chance numerically, calculate probabilities when outcomes are equally likely, and find missing probabilities. FP2 and HP2 introduce fairness assumptions; FP3 and HP3 give the probability scale; FP4 and HP4 explain why complete, non-overlapping possibilities have probabilities totalling one.

Probability scale

A bag contains three colours of counter. Must the chance of each colour be 13\frac13? Counting the colour names is not enough: we need to know how many counters carry each colour.

Probability

The probability of an event is a number describing its chance of happening. In the finite experiments here, 00 means impossible and 11 means certain. Every probability lies between 00 and 11, inclusive.

An outcome is one possible result of an experiment. An event is a collection of outcomes of interest, such as getting an even number on a die.

Probability can be written as a fraction, decimal or percentage: 12=0.5=50%\frac12=0.5=50\%. An even chance is exactly 12\frac12. A probability between 00 and 12\frac12 is unlikely; one between 12\frac12 and 11 is likely. “Likely” still allows the event not to happen.

[DIAGRAM: asset_name: The probability scale; asset_slug: probability-scale; caption: Probability runs from impossible at zero to certain at one.; description: An exact number line from zero to one marks one quarter, one half and three quarters, with impossible, unlikely, even chance, likely and certain positioned appropriately.]
Probability runs from impossible at zero to certain at one.

Worked example 1: below an even chance

Question. An event has probability 0.240.24. Is it impossible, unlikely, an even chance, likely or certain?

Plan. Compare 0.240.24 with the scale's boundary values.

0<0.24<0.50<0.24<0.5

Answer. Unlikely.

Check. It is 24%24\%, below 50%50\% but above 0%0\%, so neither an even chance nor impossible.

Change only the probability. It now lies on the other side of an even chance.

Worked example 2: above an even chance

Question. An event has probability 0.740.74. Describe its likelihood.

Plan. Locate 0.740.74 between 12\frac12 and 11.

0.5<0.74<10.5<0.74<1

Answer. Likely, but not certain.

Check. The event can still fail: its probability is less than 11.

Words describing chance must agree with these numerical limits.

Watch out

A probability of 1.21.2, or −0.2-0.2, is invalid. A probability of 20%20\% is valid because it means 0.20.2, not 2020.

Choose a description using the numerical scale.

Equally likely outcomes

If every individual outcome has the same chance, count the favourable outcomes and divide by the total. First check that the equal-chance assumption is justified.

Equally likely outcomes

P(event)=number of favourable outcomestotal number of possible outcomesP(\text{event})= \frac{\text{number of favourable outcomes}}{\text{total number of possible outcomes}}

This counting rule requires the counted outcomes to be equally likely.

Not in the formula booklet: learn it.

A fair six-sided die has each face probability 16\frac16. Choosing one counter uniformly at random means every individual counter has the same chance. Different colours need not have the same chance, because a colour can occur on several counters.

Worked example 3: two colours

Question. A bag contains 44 red and 66 blue counters. One counter is chosen uniformly at random. Find the probability that it is red.

Plan. Count individual red counters and all individual counters.

P(red)=44+6=410=25P(\text{red})=\frac4{4+6}=\frac4{10}=\frac25

Answer. 25\frac25.

Check. Four of the ten equally likely counters are red, giving 40%40\%. Having two colour names does not make the chance 12\frac12.

Split the blue group into two colours without changing the red counters or total.

Worked example 4: three colours

Question. A bag contains 44 red, 33 blue and 33 green counters. One counter is chosen uniformly at random. Find the probability that it is red.

Plan. Use the same individual-counter model, despite the extra colour name.

P(red)=44+3+3=410=25P(\text{red})=\frac4{4+3+3}=\frac4{10}=\frac25

Answer. 25\frac25.

Check. There are still four red counters among ten; changing non-red labels cannot change the red probability.

Continue this lesson.

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