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Chapter 10: Problem 64

Explain how to find and probabilities with independent events. Give an example.

Short Answer

Expert verified
The probability of independent events is calculated by multiplying the probabilities of each individual event. For instance, if the probability of getting heads in a coin flip is \( \frac{1}{2} \), and the probability of rolling a 3 in a six-sided dice is \( \frac{1}{6} \), then the probability of both these events occurring is \( \frac{1}{2} \times \frac{1}{6} = \frac{1}{12} \).

Step by step solution

01

Define Independent Events

Independent events are those events whose occurrence or non-occurence does not affect the probability of the occurrence of the other event. In other words, the outcome of one event does not influence the result of the second event. For instance, when flipping a coin, the outcome of the first flip does not affect the outcome of the second flip.
02

Find the Probability of Each Event

In order to calculate the probability of independent events, first find the probability of each individual event. The individual event's probability is calculated by dividing the number of successful outcomes by the number of total outcomes. For instance, when flipping a coin, the probability of getting heads (successful outcome) is \( \frac{1}{2} \), as there are 2 possible outcomes (heads and tails), and only one of them is heads. Similarly, the probability of rolling a 3 on a 6-sided dice (another individual event) would be \( \frac{1}{6} \), as there are 6 outcomes, and only one of them is 3.
03

Multiply the Individual Probabilities

Once the probability of each event is determined, the next step is to multiply these probabilities in order to get the combined probability of the independent events. Let's extend the examples above. If the question is about the probability of flipping a coin and getting heads and then rolling a dice and getting a 3, this would be the product of the individual probabilities: \( \frac{1}{2} \) (for the coin) times \( \frac{1}{6} \) (for the dice), which equals \( \frac{1}{12} \). This means there's 1 in 12 chances to get heads in the coin and 3 on the dice, if both are done simultaneously.

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Most popular questions from this chapter

Will help you prepare for the material covered in the next section. Each exercise involves observing a pattern in the expanded form of the binomial expression \((a+b)^{n}\). $$\begin{array}{l} (a+b)^{1}=a+b \\ (a+b)^{2}=a^{2}+2 a b+b^{2} \\ (a+b)^{3}=a^{3}+3 a^{2} b+3 a b^{2}+b^{3} \\ (a+b)^{4}=a^{4}+4 a^{3} b+6 a^{2} b^{2}+4 a b^{3}+b^{4} \\ (a+b)^{5}=a^{5}+5 a^{4} b+10 a^{3} b^{2}+10 a^{2} b^{3}+5 a b^{4}+b^{5} \end{array}$$ Describe the pattern for the exponents on \(b\)

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Use mathematical induction to prove that each statement is true for every positive integer \(n\). $$(a b)^{n}=a^{n} b^{n}$$

Graph the piecewise function: $$ f(x)=\left\\{\begin{array}{lll} 2 x-4 & \text { if } & x \neq 3 \\ -5 & \text { if } & x=3 \end{array}\right. $$

Explain how to find or probabilities with mutually exclusive events. Give an example.
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