
What is product rule in probability?
Answer
455.1k+ views
Hint: As we know that probability is the branch of mathematics which tells us how likely an event is to occur, or how likely it is that a proposition is true. The probability of an event is a number between $ 0 $ and $ 1 $ where $ 0 $ indicates impossibility of the event and $ 1 $ indicates the possibility of the event. To solve the questions of probability we introduced many rules.
Complete step by step solution:
One of the basic rules of probability that`s very useful to solve the questions is product rule. The product rule states that the probability of two (or more) independent events occurring together can be calculated by multiplying the individual probabilities of the event. Specifically, the rule of product is used to find the probability of an intersection of events:
Let $ A $ and $ B $ be independent events. Then,
$ \Rightarrow P\left( A\cap B \right)=P\left( A \right)\times P\left( B \right) $
Let take a familiar example to understand the concept of product rule:
Example- A fair die is rolled twice then what is the probability that both rolls have a result of $ 3 $ ?
Solution- Here the die rolled independently, that is if the first die roll it will not affect the probability of the second die roll resulting in $ 3 $ .
Let probability of getting 3 when die is roll once is $ A $ and probability of getting $ 3 $ when die is roll second time is $ B $
The probability of rolling $ 3 $ for an individual die roll is $ \dfrac{1}{6} $ . Now here we will apply the product rule because it is asking the probability of $ 3 $ when we roll die twice.
So this problem is asking for $ P $ (1st roll is $ 3\cap $ 2nd roll is $ 3 $ )
Therefore by using product rule, we get
$ \begin{align}
& \Rightarrow P\left( A\cap B \right)=P\left( A \right)\times P\left( B \right) \\
& \Rightarrow P\left( A\cap B \right)=\dfrac{1}{6}\times \dfrac{1}{6} \\
& \Rightarrow P\left( A\cap B \right)=\dfrac{1}{36} \\
\end{align} $
Hence the probability of getting $ 3 $ when both die roll is $ \dfrac{1}{36} $ .
Note: It is important to note that the product rule is used when events are independent. If we have dependent events then we will use another rule. The probability of an intersection of dependent events then we use a different approach involving conditional probability.
Complete step by step solution:
One of the basic rules of probability that`s very useful to solve the questions is product rule. The product rule states that the probability of two (or more) independent events occurring together can be calculated by multiplying the individual probabilities of the event. Specifically, the rule of product is used to find the probability of an intersection of events:
Let $ A $ and $ B $ be independent events. Then,
$ \Rightarrow P\left( A\cap B \right)=P\left( A \right)\times P\left( B \right) $
Let take a familiar example to understand the concept of product rule:
Example- A fair die is rolled twice then what is the probability that both rolls have a result of $ 3 $ ?
Solution- Here the die rolled independently, that is if the first die roll it will not affect the probability of the second die roll resulting in $ 3 $ .
Let probability of getting 3 when die is roll once is $ A $ and probability of getting $ 3 $ when die is roll second time is $ B $
The probability of rolling $ 3 $ for an individual die roll is $ \dfrac{1}{6} $ . Now here we will apply the product rule because it is asking the probability of $ 3 $ when we roll die twice.
So this problem is asking for $ P $ (1st roll is $ 3\cap $ 2nd roll is $ 3 $ )
Therefore by using product rule, we get
$ \begin{align}
& \Rightarrow P\left( A\cap B \right)=P\left( A \right)\times P\left( B \right) \\
& \Rightarrow P\left( A\cap B \right)=\dfrac{1}{6}\times \dfrac{1}{6} \\
& \Rightarrow P\left( A\cap B \right)=\dfrac{1}{36} \\
\end{align} $
Hence the probability of getting $ 3 $ when both die roll is $ \dfrac{1}{36} $ .
Note: It is important to note that the product rule is used when events are independent. If we have dependent events then we will use another rule. The probability of an intersection of dependent events then we use a different approach involving conditional probability.
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