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A pair of dice is rolled, and the number that appears uppermost on each die is observed.refer to this experiment and find the probability of the given event. A double is thrown.

Short Answer

Expert verified
The probability of rolling a double when throwing a pair of dice is \(\frac{1}{6}\).

Step by step solution

01

Identify successful outcomes

First, we need to determine the successful outcomes for rolling a double. These are the combinations where both dice show the same number: (1,1), (2,2), (3,3), (4,4), (5,5), and (6,6).
02

Calculate total possible outcomes

Next, identify the total possible outcomes when rolling two dice. Since each die has 6 faces, there are 6 * 6 = 36 possible outcomes.
03

Calculate the probability

Finally, to find the probability of rolling a double, we divide the number of successful outcomes by the total number of possible outcomes. Probability of rolling a double = (Number of successful outcomes) / (Total number of possible outcomes) = \(\frac{6}{36}\)
04

Simplify the probability

Now, simplify the fraction by dividing both the numerator and denominator by their greatest common divisor, which is 6. \(\frac{6}{36}\) = \(\frac{6 \div 6}{36 \div 6}\) = \(\frac{1}{6}\) So, the probability of rolling a double when throwing a pair of dice is \(\frac{1}{6}\).

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Key Concepts

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

Dice outcomes
When you roll two dice, each die can land on a number from 1 to 6. This means there is a wide range of combinations that can appear as outcomes. Each die acts independently, so the outcome on one die does not affect the other.
Given two dice, which we can label as Die A and Die B, each die when rolled has possibilities like:
  • Die A: 1, 2, 3, 4, 5, or 6
  • Die B: 1, 2, 3, 4, 5, or 6
The total number of outcomes when rolling two dice is calculated by multiplying the outcomes for Die A and Die B, so 6 outcomes from Die A times 6 outcomes from Die B result in 36 total possible outcomes.
This means that with two dice, you have 36 possible pairs of results that can appear on the top faces.
Probability calculation
Probability is about finding how likely an event is to happen. To calculate probability, we need two key pieces of information:
  • The number of desired or successful outcomes
  • The total number of possible outcomes
For rolling doubles with two dice, successful outcomes are when both dice show the same number. As identified earlier, these are: (1,1), (2,2), (3,3), (4,4), (5,5), and (6,6). This gives us 6 successful outcomes.
The probability formula therefore can be set up as:\[ \text{Probability of rolling a double} = \frac{\text{Number of successful outcomes}}{\text{Total number of possible outcomes}} \]Inserting our numbers into the formula gives:\(\frac{6}{36}\).
This formula helps us recognize and calculate how frequent rolling doubles can be expected from two dice.
Simplifying fractions
Fractions often need to be simplified to make them easier to understand or compare. Simplification involves reducing the numerator (top number) and the denominator (bottom number) to their smallest possible whole numbers.
In the fraction \(\frac{6}{36}\), both the numerator and denominator can be divided by their greatest common divisor (GCD). The GCD here is 6, which is the largest number that evenly divides both 6 and 36.
So, we divide the numerator and the denominator by 6:\[ \frac{6 \div 6}{36 \div 6} = \frac{1}{6} \]
By simplifying \(\frac{6}{36}\) to \(\frac{1}{6}\), we can easily understand that the probability of rolling the same number on both dice (a double) is 1 in 6 tries.

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Most popular questions from this chapter

In "The Numbers Game," a state lottery, four numbers are drawn with replacement from an urn containing balls numbered \(0-9\), inclusive. Find the probability that a ticket holder has the indicated winning ticket. Three digits in exact order

The following table summarizes the results of a poll conducted with 1154 adults. $$\begin{array}{lcccc} \text { Income, \$ } & \text { Range, \% } & \text { Rich, \% } & \text { Class, \% } & \text { Poor, \% } \\ \hline \text { Less than 15,000 } & 11.2 & 0 & 24 & 76 \\ \hline 15,000-29,999 & 18.6 & 3 & 60 & 37 \\ \hline 30,000-49,999 & 24.5 & 0 & 86 & 14 \\ \hline 50,000-74,999 & 21.9 & 2 & 90 & 8 \\ \hline 75,000 \text { and higher } & 23.8 & 5 & 91 & 4 \\ \hline \end{array}$$ a. What is the probability that a respondent chosen at random calls himself or herself middle class? b. If a randomly chosen respondent calls himself or herself middle class, what is the probability that the annual household income of that individual is between $$\$ 30.000$$ and $$\$ 49,999$$, inclusive? c. If a randomly chosen respondent calls himself or herself middle class, what is the probability that the individual's income is either less than or equal to $$\$ 29,999$$ or greater than or equal to $$\$ 50,000$$ ?

If a 5-card poker hand is dealt from a well-shuffled deck of 52 cards, what is the probability of being dealt the given hand? A straight flush (Note that an ace may be played as either a high or low card in a straight sequence- that is, \(\mathrm{A}, 2,3\), 4,5 or \(10, \mathrm{~J}, \mathrm{Q}, \mathrm{K}, \mathrm{A}\). Hence, there are ten possible sequences for a straight in one suit.)

Copykwik has four photocopy machines: \(A, B, C\), and \(D .\) The probability that a given machine will break down on a particular day is \(P(A)=\frac{1}{50} \quad P(B)=\frac{1}{60} \quad P(C)=\frac{1}{75} \quad P(D)=\frac{1}{40}\) Assuming independence, what is the probability on a particular day that a. All four machines will break down? b. None of the machines will break down?

The chief loan officer of La Crosse Home Mortgage Company summarized the housing loans extended by the company in 2007 according to type and term of the loan. Her list shows that \(70 \%\) of the loans were fixed-rate mortgages \((F), 25 \%\) were adjustable-rate mortgages \((A)\), and \(5 \%\) belong to some other category \((O)\) (mostly second trust-deed loans and loans extended under the graduated payment plan). Of the fixed-rate mortgages, \(80 \%\) were 30 -yr loans and \(20 \%\) were 15 -yr loans; of the adjustable-rate mortgages, \(40 \%\) were 30 -yr loans and \(60 \%\) were 15 -yr loans; finally, of the other loans extended, \(30 \%\) were 20 -yr loans, \(60 \%\) were 10 -yr loans, and \(10 \%\) were for a term of 5 yr or less. a. Draw a tree diagram representing these data. b. What is the probability that a home loan extended by La Crosse has an adjustable rate and is for a term of 15 yr? c. What is the probability that a home loan cxtended by La Crosse is for a term of 15 yr?

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