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(a) Suppose that you throw 4 dice. Find the probability that you get at least one 1 . (b) Suppose that you throw 2 dice 24 times. Find the probability that you get at least one \((1,1),\) that is, you roll "snake;-eyes." [Note: The probability of part (a) is greater than the probability of part (b).]

Short Answer

Expert verified
(a) The probability of getting at least one '1' in four throws of a dice is \(1 - (5/6)^4\). (b) The probability of getting at least one pair of '1,1' in 24 throws of a pair of dice is \(1 - (35/36)^{24}\).

Step by step solution

01

Understand and set up the problem for Part (a)

The problem involves four dice being thrown. The goal is to find the probability of getting at least one '1'. The opposite (or the complement) of getting at least one '1' is not getting any '1's at all. The problem can be solved quicker by finding the probability of this complementary event and using it to determine the desired probability.
02

Calculate the probability of complementary event for Part (a)

The probability of not getting a '1' on a single dice throw is \(5/6\) (as there are five other possible outcomes: 2,3,4,5 and 6). Since the dice are assumed to be fair and the throws are independent, the probability of not getting any '1's in four throws is \((5/6)^4\).
03

Calculate the desired probability from the complementary event for Part (a)

The desired probability is simply 1 minus the probability of the complementary event. Therefore, the probability of getting at least one '1' in four throws is \(1 - (5/6)^4\).
04

Understand and set up the problem for Part (b)

The problem involves two dice being thrown 24 times. The goal is to find the probability of getting at least one pair of '1,1'. As in Part (a), the opposite (or the complement) of getting at least one pair of '1,1' is not getting any pairs of '1,1'. The problem again can be solved quicker by finding the probability of this complementary event and using it to determine the desired probability.
05

Calculate the probability of complementary event for Part (b)

The probability of not getting a pair of '1,1' on a single pair of dice throws is \(35/36\) (as there are 36 possible outcomes and only one of them is '1,1'). The probability of not getting any pairs of '1,1' in 24 throws is \((35/36)^{24}\).
06

Calculate the desired probability from the complementary event for Part (b)

Again, the desired probability is simply 1 minus the probability of the complementary event. Therefore, the probability of getting at least one pair of '1,1' in 24 throws is \(1 - (35/36)^{24}\).

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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 that deals with counting, arranging, and combining items. It's very useful in probability theory because it helps us understand how many ways an event can occur. In the context of the exercise, combinatorics helps to determine possible outcomes when dice are rolled. For example, when rolling a single die, there are 6 possible outcomes. When two dice are rolled together, the number of potential combinations is 36 (since each die has 6 faces, and you multiply them together: 6 × 6).

In general, combinatorics provides tools to calculate the number of ways certain things can occur, which is crucial for computing probabilities. Understanding these concepts can make it easier to predict the likelihood of different events, such as calculating how often snake eyes appear when rolling two dice multiple times.
Independent Events
Independent events are events where the outcome of one event does not affect the outcome of another. This is an important concept in probability, particularly when you're dealing with scenarios like rolling dice.

For example, if you roll a die multiple times, the result of any single roll doesn't influence the results of the others. Each die roll stands alone, with the same probability of outcomes each time.
  • The probability of rolling a specific number, such as a '1', on one roll remains \( \frac{1}{6} \) regardless of previous rolls.
  • When calculating probabilities of outcomes involving multiple rolls, you multiply individual probabilities, such as finding the probability of not rolling a '1' over several throws, as seen in the exercise.
Understanding that events are independent allows for using simple multiplication to find the probability over multiple attempts.
Complementary Events
Complementary events are pairs of events that are mutually exclusive and encompass all possibilities. If one event occurs, the other cannot, and together, their probabilities add up to 1.

In our exercise, the events 'rolling at least one 1' and 'rolling no 1s at all' are complements of each other. The probability of experiencing one event is the complement of the probability of experiencing the other.
  • For instance, if the probability of rolling no 1s with four dice is \( \left( \frac{5}{6} \right)^4 \), then the probability of getting at least one 1 is 1 minus this probability: \( 1 - \left( \frac{5}{6} \right)^4 \).
  • This approach simplifies calculations by focusing on the complement, which is often easier to compute.
Using complementary events is a strategic tool in probability theory to simplify complex problems.
Probability Distribution
In probability theory, a probability distribution describes how probabilities are distributed over the values of a random variable. It tells us how the outcomes of an event, like dice throws, are spread out.

For example, when throwing two dice, the probability distribution shows how likely each possible sum outcome (from 2 to 12) is. Each pair of results from the two dice has a probability of occurring, and the total event probabilities must sum to 1.
  • In exercises, understanding probability distributions allows you to calculate specific probabilities, like the distribution for rolling 'snake eyes' (a 1 on each die).
  • The more we repeat the experiment (like 24 dice throws), the more the probability distribution becomes significant in predicting outcomes over multiple events.
Probability distributions are fundamental for predicting overall behaviors of random processes, such as gambling, genetics, or any systems where outcomes are unpredictable.

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