Independent events and the multiplication rule ·… | StemPy Lab
Course 3.2·Lesson 1 of 3
Independent events and the multiplication rule
Two events are independent when one of them happening does not change the chance of the other. Then the chance that both happen is the product, P(A∩B)=P(A)×P(B)
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Try the code and the simulations beside the lesson.
What you will be able to do
decide whether two events can be treated as independent;
use the multiplication rule P(A∩B)=P(A)×P(B) for independent events;
find the chance of "at least one" as 1−P(none).
A coin and a die
Toss a fair coin and roll a fair die. What is the chance of a head and a six?
List the outcomes. The coin has 2 faces and the die has 6, so there are 2×6=12 equally likely pairs. Only one of them is a head with a six. So P(head and six)=121.
A grid with two rows and six columns showing the 12 equally likely results of a coin and a die. The rows are head and tail, and the columns are the die faces 1 to 6. One cell, a head with a six, is shaded orange, and the top row of six cells with a head has a blue border.
Now look at it another way. Half of the pairs have a head. Among those, one in six has a six. So the answer is 21×61=121. The coin does not care what the die does, and the die does not care about the coin.
Independent events
Two events are independent when knowing that one happened does not change the probability of the other. Then
P(A∩B)=P(A)×P(B)
This is the multiplication rule for independent events. It works for more events too: for three independent events, multiply three probabilities.
Some pairs are clearly independent: two different coins, or a coin and a die. Many real pairs are not.
Rain today and rain tomorrow in one town. A wet day is often followed by another wet day.
Two cards from one pack without putting the first back. The first card changes what is left.
Two events about the same die. "Even" and "more than 3" share the outcomes 4 and 6.
For the last pair, P(even)=21 and P(more than 3)=21. Their product is 41. But the real chance of both is 62=31. The product rule gives the wrong answer, so these events are not independent.
You meet the exam form of independence, with the test P(A∣B)=P(A), in the Maths course on conditional probability and independent events.
Try it in Python
The program rolls a computer die and tosses a computer coin 60000 times. It compares the share of "head and six" with the product 21×61. Then it does the same for "even and more than 3" on one die.
At least one
Two made-up alarms guard a door, and they work independently. Each rings in a break-in with probability 0.9. What is the chance that at least one rings?
"At least one" covers several cases: the first only, the second only, or both. It is quicker to find the only case left out, "none":
P(none)=0.1×0.1=0.01
P(at least one)=1−0.01=0.99
The same idea works for many tries. An event with probability 0.4, tried 3 times independently, happens at least once with probability 1−0.63=1−0.216=0.784.
Your turn
The usual mistakes
Multiplying when the events are linked. The product rule needs independence. Ask whether one event changes the chances of the other.
Adding for "and". For independent events, "and" multiplies. Adding can even give a number above 1.
Mixing up independent and mutually exclusive. Exclusive events cannot happen together, so knowing one happened tells you a lot about the other. With positive chances they are never independent.
Counting the cases of "at least one" one by one. Use 1−P(none).
Remember these
Independent: one event does not change the chance of the other.
Independent events: P(A∩B)=P(A)×P(B).
P(at least one)=1−P(none).
Check yourself
Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.
A fair coin is tossed and a fair die is rolled. Find the probability of a tail and an odd number.
Are "a student is in the school choir" and "the same student's twin is in the school choir" likely to be independent? Why?
Three independent made-up machines each work on a given day with probability 0.9. Find the probability that all three work.
Each ticket in a made-up game wins with probability 0.2, independently. Find the probability of at least one win with 3 tickets.
For two events, P(A)=0.5, P(B)=0.4 and P(A∩B)=0.3. Are they independent?
References
OpenStax, Introductory Statistics, Chapter 3, the section on independent and mutually exclusive events.
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