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Chapter 10: Problem 34

Two dice are rolled. Find the probability that at least one of the dice is a one.

Short Answer

Expert verified
The probability is \(\frac{11}{36}\).

Step by step solution

01

Understand the Sample Space

When two dice are rolled, each die has 6 faces. Therefore, there are a total of 6 * 6 = 36 possible outcomes (ordered pairs).
02

Identify the Favorable Outcomes

Count the outcomes where at least one die shows a one. These can be listed as: \[(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (3,1), (4,1), (5,1), (6,1)\]. There are 11 outcomes where at least one die shows a one.
03

Calculate the Probability

Probability is calculated as the number of favorable outcomes divided by the total number of possible outcomes. Therefore: \[\text{Probability} = \frac{11}{36}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

sample space in probability
In probability, the sample space is the set of all possible outcomes of an experiment. For example, when rolling two dice, we need to consider every possible result. Each die has 6 faces, thus for two dice, we calculate the total outcomes by multiplying 6 by 6, giving us 36 possible ordered pairs. These outcomes include pairs like (1,1), (1,2), and continue up to (6,6). This comprehensive list is known as the sample space.
favorable outcomes
Favorable outcomes are specific results from the sample space that meet the condition we are interested in. For our problem, we are looking for outcomes where at least one of the dice shows a one. We list these outcomes as follows: (1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (3,1), (4,1), (5,1), and (6,1). These 11 pairs are our favorable outcomes since they satisfy the requirement of having at least one die showing a one.
calculating probability
To calculate the probability of an event, we use the ratio of the number of favorable outcomes to the total number of possible outcomes. The formula is \(\text{Probability} = \frac{\text{Number of Favorable Outcomes}}{\text{Total Number of Possible Outcomes}}\). In our example, the number of favorable outcomes is 11, and the total number of possible outcomes is 36. Therefore, the probability that at least one of the dice shows a one is \(\frac{11}{36}\). This fraction represents the likelihood of our desired event occurring.

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Most popular questions from this chapter

Cavities in trees are necessary for certain birds and other animals to survive. For a plot of woods in Missouri, let \(X\) represent the number of cavities per hectare. The density function for \(X\) can be approximated by \(^{24}\) $$ f(x)=\frac{a}{b}\left(\frac{x-c}{b}\right)^{a-1} e^{-(x-c)^{a} b^{a}} $$ where \(a, b\), and \(c\) are constants, and \(x \geq c\). a) Verify property 2 of a density function for \(\int(x)\). (Hint: Try substituting \(u=(x-c) / b .)\) b) For plots of trees no more than 30 yr old, it was found that \(a \approx 0.68, b \approx 0.89\), and \(c \approx 0.0\) Compute \(P(X \leq 2)\).

A random variable \(X\) is assumed to have a standard normal distribution. Find the observed significance level if the random variable is equal to the given value in an experiment. Observed value of \(-1.24\)

Compute the probability using the given distribution. \(P(10 \leq X \leq 30), X\) is Uniform \((-10,70)\)

Use the definition of a density function to verify that $$ f(x)=\lambda e^{-\lambda x}, \quad \text { for } x \geq 0, $$ is a probability density function for any positive value of \(\lambda\).

Assume that the random variable \(X\) is normally distributed. Use the given information to find the unknown parameter or parameters of the distribution. If \(S D(X)=3\) and \(P(X \geq 2)=0.6293\), find \(E(X)\)
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