Probability - Rule of Product
The rule of product is a guideline as to when probabilities can be multiplied to produce another meaningful probability. Specifically, the rule of product is used to find the probability of an intersection of events:
Let \(A\) and \(B\) be independent events. Then,
\(P(A\cap B)=P(A)\times P(B)\)
A 6-sided fair die is rolled twice. What is the probability that both rolls have a result of 6?
It is important to establish that each die roll is independent. That is, if the first die roll result is 6, it will not affect the probability of the second die roll resulting in 6.
For an individual die roll, the probability of rolling 6 is \(\dfrac{1}{6}\).
Effectively, this problem is asking for \(P(\text{1st roll is 6}\cap\text{2nd roll is 6})\).
Using the rule of product, this is:
\(\dfrac{1}{6}\times\dfrac{1}{6}=\dfrac{1}{36}\).
The probability that both die rolls are 6 is \(\boxed{\dfrac{1}{36}}\).
An important requirement of the rule of product is that the events are independent. If one were to calculate the probability of an intersection of dependent events, then a different approach involving conditional probability would be needed.
Rule of Product for Independent Events
Calvin has \(5\) different ties, \(6\) different shirts, and \(4\) different pants to wear in his wardrobe. Of those clothing items, Calvin has exactly one green tie, exactly one gray shirt, and exactly one pair of black pants. If Calvin randomly selects each item of clothing at random, what is the probability that he will wear the green tie, gray shirt, and black pants?
There is a \(\dfrac{1}{5}\) probability that Calvin would randomly select the green tie, a \(\dfrac{1}{6}\) probability that he would randomly select the gray shirt, and a \(\dfrac{1}{4}\) probability that he would randomly select the black pants.
These events are independent; the selection of the tie does not affect the selection of the shirt, and so on.
Hence, the probability that he selects the green tie, gray shirt, and black pants is \(\dfrac{1}{5} \times \dfrac{1}{6} \times \dfrac{1}{4} = \boxed{\dfrac{1}{120}}\).
You have three 6-sided dice. One is red, one is blue, and one is green. What is the probability that you will roll a 1 on the red die, a 2 on the blue die, and a 3 on the green die?
There are 5 possible routes (red, blue, green, yellow, purple) from the Candy Castle to the Chocolate Forest, and 3 routes (pink, white, orange) from the Chocolate Forest to the Ice Cream Cottage, if each path is chosen at random, what is the probability you will travel from the Candy Castle to the Ice Cream Cottage via the purple and orange road?
There is a \(\dfrac{1}{5}\) chance of choosing the purple path, and there is a \(\dfrac{1}{3}\) chance of choosing the orange path. The selection of the path to the Chocolate Forest is independent of the selection of the path to the Ice Cream Cottage.
By the rule of product, there is a \(\dfrac{1}{5} \times \dfrac{1}{3} = \boxed{\dfrac{1}{15}}\) chance of choosing the purple-orange path.
There are 4 human friends of the Mystery Inc. (assuming Scooby-Doo is always with Shaggy) running away from a mysterious ghost, and there are 4 available rooms to hide in.
Any of these rooms can contain up to 4 teenagers (or none at all).
If the ghost opens one of these 4 rooms in random, what is the probability that he will see no-one hiding in there?
If this probability can be expressed as \(\dfrac{a}{b}\), where \(a\) and \(b\) are coprime positive integers, enter \(a+b\) as your answer.
Rule of Product for Dependent Events
When events are dependent, it can be more challenging to calculate the probability of the intersection of those events. This calculation requires the knowledge of a conditional probability.
Let \(A\) and \(B\) be dependent events. Then,
\(P(A\cap B)=P(A)\times P(B\mid A)\)
A bug stands on the topmost vertex of the pentagonal trapezohedron pictured below.
Every 10 seconds, the bug selects an adjacent edge at random, and moves along it to another vertex. What is the probability that the bug will be on the bottom vertex after 30 seconds?
Image Credit: Mouagip
The bug has 5 adjacent edges to choose from in the first 10 seconds, but it doesn't matter which one he chooses, as they all bring him closer to the bottom. The probability that the bug gets closer to the bottom after the first 10 seconds is \(1\).
The bug has 3 adjacent edges to choose from in the next 10 seconds. 2 of those edges bring him closer to the bottom. The probability that the bug gets closer to the bottom after these 10 seconds is \(\dfrac{2}{3}\).
Given that the bug moved closer after the previous 10 seconds, the bug has 3 adjacent edges to choose from in the final 10 seconds. 1 of those edges will bring the bug to the bottom. Given that the bug moved correctly previously, the probability that the bug makes it to the bottom after the final 10 seconds is \(\dfrac{1}{3}\).
Using the product rule for dependent events, the probability that the bug makes it to the bottom after 30 seconds is \(1\times \dfrac{2}{3}\times\dfrac{1}{3}=\boxed{\dfrac{2}{9}}\)
A bug starts on a vertex and randomly moves along one edge of an icosahedron every 10 seconds to another vertex. What is the probability that it will end up at the opposite vertex after 30 seconds?
If the probability is expressed as \( \dfrac ab\), where \(a\) and \(b\) are coprime positive integers, find \(a+b\).
Try more questions on Platonic Solids.
Image credit: http://home.btconnect.com/