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In a single throw of two dice, what is the probability of getting (a) a total of 5 , (b) a total of at most 5 , (c) a total of at least 5 ?

Short Answer

Expert verified
(a) \(\frac{1}{9}\), (b) \(\frac{5}{18}\), (c) \(\frac{13}{18}\)

Step by step solution

01

Understanding the Problem

In this exercise, we need to determine the probability of outcomes when rolling two dice. Each die can show a number from 1 to 6. With two dice, the total outcomes are 6 (from the first die) multiplied by 6 (from the second die), which equals 36 possible outcomes.
02

Calculating Probability of Total 5

List all the combinations that sum up to 5:- Die 1 shows 1 and Die 2 shows 4: (1,4)- Die 1 shows 2 and Die 2 shows 3: (2,3)- Die 1 shows 3 and Die 2 shows 2: (3,2)- Die 1 shows 4 and Die 2 shows 1: (4,1)There are 4 successful outcomes. Probability = \(\frac{4}{36} = \frac{1}{9}\).
03

Calculating Probability of Total at Most 5

"At most 5" means the total can be 2, 3, 4, or 5. List the combinations:- Total 2: (1,1)- Total 3: (1,2), (2,1)- Total 4: (1,3), (2,2), (3,1)- Total 5: (1,4), (2,3), (3,2), (4,1)Count the combinations: 1 + 2 + 3 + 4 = 10. Probability = \(\frac{10}{36} = \frac{5}{18}\).
04

Calculating Probability of Total at Least 5

"At least 5" means the total can be 5, 6, 7, 8, 9, 10, 11, or 12. Calculate by complementary probability: Total outcomes are 36. Outcomes fewer than 5 are 10 (from Step 3). Outcomes at least 5 = 36 - 10 = 26. Probability = \(\frac{26}{36} = \frac{13}{18}\).

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Key Concepts

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

Combinatorics
Combinatorics is a branch of mathematics concerned with counting, arranging, and finding patterns. It helps in understanding how different combinations are formed. When considering a problem with multiple events, like rolling dice, combinatorics helps in calculating the possible outcomes.

In the context of rolling two dice, each die has 6 faces, resulting in a variety of combinations when rolled together. The principle of counting states that each face of the first die can pair with each face of the second die. This gives us a total of 6 times 6, equaling 36 possible outcomes. Understanding this is crucial, as it forms the basis for calculating probabilities later on.

By grasping combinatorics, you'll know how to systematically list or compute all outcomes before calculating the probability of specific events, ensuring you don't miss any potential result.
Dice outcomes
Dice are popular tools for learning probability and combinatorics. Through two standard six-sided dice, we can explore various potential outcomes.

Each die displays a number from 1 to 6, yielding 36 outcomes when two dice are rolled. These outcomes allow us to calculate the probability of different sums appearing. For instance, to find the combinations that add up to a specific number, such as 5, list the pairs of dice that yield the desired sum:
  • (1,4)
  • (2,3)
  • (3,2)
  • (4,1)
This list reveals 4 combinations resulting in a total of 5.

Listing all possible outcomes makes it simpler to determine the likelihood of various totals. Thus, mastering the concept of dice outcomes enhances our ability to approach more complicated probability challenges.
Probability calculations
Probability calculations involve determining how likely an event is to occur within a set frame of outcomes. It's expressed as a fraction or percentage.

In the realm of dice, the probability is found by dividing the favorable outcomes by the total possible outcomes. For example, to calculate the probability of rolling a sum of 5, you'll use the 4 favorable outcomes obtained from our list of dice pairs. With the total outcomes being 36, the probability equals \(\frac{4}{36}\) or \(\frac{1}{9}\).

Let's consider the probability of rolling a sum of "at most 5." We add the numbers from totals 2, 3, 4, and 5 to get 10 successful outcomes. Hence, the probability becomes \(\frac{10}{36}\) or \(\frac{5}{18}\).

Lastly, understanding complementary probability helps when dealing with totals "at least 5." Instead of listing every combination from 5 to 12, identify the easier route by subtracting totals for sums lower than 5 from the total possible outcomes. That results in 26 successful outcomes and hence a probability of \(\frac{26}{36}\) or \(\frac{13}{18}\).

Probability calculations demystify how often you'd expect each event to occur in practice, providing a basis for gamified learning and critical thinking.

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Most popular questions from this chapter

Suppose that you roll a pair of ordinary dice repeatedly until you get either a total of seven or a total of ten. What is the probability that the total then is seven?

The diameter of ball-bearings produced by a machine is a random variable having a normal distribution with mean \(6.00 \mathrm{~mm}\) and standard deviation \(0.02 \mathrm{~mm}\). If the diameter tolerance is \(\pm 1 \%\), find the proportion of ball-bearings produced that are out of tolerance. After several years' use, machine wear has the effect of increasing the standard deviation, although the mean diameter remains constant. The manufacturer decides to replace the machine when \(2 \%\) of its output is out of tolerance. What is the standard deviation when this happens?

Suppose that a coin is tossed three times and that the random variable \(W\) represents the number of heads minus the number of tails. (a) List the elements of the sample space \(S\) for the three tosses of the coin, and to each sample point assign a value \(w\) of \(W\). (b) Find the probability distribution of \(W\), assuming that the coin is fair. (c) Find the probability distribution of \(W\), assuming that the coin is biased so that a head is twice as likely to occur as a tail.

The 'odds' in favour of an event \(A\) are quoted as \(' a\) to \(b\) ' if and only if \(P(A)=a /(a+b)\). The 'odds against' are then ' \(b\) to \(a\) ' (which is the usual way to quote odds in betting situations). (a) If an insurance company quotes odds of 3 to 1 in favour of an individual 70 years of age surviving another ten years, what is the corresponding probability? (b) If the probability of a successful transplant operation is \(\frac{1}{8}\), what are the odds against success?

A continuous random variable \(X\) has probability density function given by $$ f_{X}(x)= \begin{cases}\frac{c}{x^{4}} & \text { for } x \geqslant 1 \\ 0 & \text { for } x<1\end{cases} $$ where \(c\) is constant. Find (a) the value of the constant \(c\); (b) the cumulative distribution function of \(X\); (c) \(P(X>2)\); (d) the mean of \(X\); (e) the standard deviation of \(X\).

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