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You roll two fair dice, a green one and a red one. (a) What is the probability of getting a sum of \(7 ?\) (b) What is the probability of getting a sum of \(11 ?\) (c) What is the probability of getting a sum of 7 or 11? Are these outcomes mutually exclusive?

Short Answer

Expert verified
(a) \( \frac{1}{6} \), (b) \( \frac{1}{18} \), (c) \( \frac{2}{9} \); Yes, they are mutually exclusive outcomes.

Step by step solution

01

Understand the total outcomes

When two dice are rolled, each die has 6 faces. Thus, the total number of outcomes when rolling two dice is calculated by multiplying the number of faces on each die: \[6 \times 6 = 36\].This means there are 36 possible outcomes when rolling two dice.
02

Identify outcomes for sum of 7

The ways to get a sum of 7 are: - Green die: 1, Red die: 6 - Green die: 2, Red die: 5 - Green die: 3, Red die: 4 - Green die: 4, Red die: 3 - Green die: 5, Red die: 2 - Green die: 6, Red die: 1 There are a total of 6 combinations that result in a sum of 7.
03

Calculate probability for sum of 7

The probability of getting a sum of 7 is the number of successful outcomes divided by the total number of outcomes:\[P(7) = \frac{6}{36} = \frac{1}{6}\].
04

Identify outcomes for sum of 11

The ways to obtain a sum of 11 are: - Green die: 5, Red die: 6 - Green die: 6, Red die: 5 There are a total of 2 combinations that result in a sum of 11.
05

Calculate probability for sum of 11

The probability of getting a sum of 11 is the number of successful outcomes divided by the total number of outcomes:\[P(11) = \frac{2}{36} = \frac{1}{18}\].
06

Calculate probability for sum of 7 or 11

The probability of getting either a sum of 7 or a sum of 11 is determined by adding their individual probabilities:\[P(7 \text{ or } 11) = P(7) + P(11) = \frac{1}{6} + \frac{1}{18}\].Bringing to a common denominator we get:\[P(7 \text{ or } 11) = \frac{3}{18} + \frac{1}{18} = \frac{4}{18} = \frac{2}{9}\].
07

Determine if outcomes are mutually exclusive

Two outcomes are mutually exclusive if they cannot happen at the same time. Since a roll of two dice cannot result in a sum of both 7 and 11 simultaneously, these two outcomes are mutually exclusive.

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Key Concepts

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

Mutually Exclusive Events
In probability, when we talk about mutually exclusive events, we refer to events that cannot occur at the same time. For example, when rolling two dice, you cannot get a sum of both 7 and 11 at the same time from a single roll.
Hence, these outcomes are considered mutually exclusive.

If two events are mutually exclusive, it means that the occurrence of one event means the other cannot occur in that instance. This is an important concept when calculating probabilities because it helps us understand how different events relate
to one another. Mutual exclusivity simplifies calculations since the joint probability of both occurring is zero.

Understanding mutually exclusive events helps pave the way for determining the probability of either event happening by using the rule:
  • If events A and B are mutually exclusive, \(P(A \text{ or } B) = P(A) + P(B)\).
Outcomes in Dice Rolls
Rolling dice is a classic example that helps illustrate various concepts in probability. When we roll two dice, each die has 6 faces, usually numbered from 1 to 6.
The total number of possible outcomes when rolling two dice is calculated by multiplying the number of outcomes for each die: \(6 \times 6 = 36\).
The outcomes of interest in dice rolls are often specific sums of the numbers appearing on each die. For instance, if you need to find the combinations that lead to specific sums, like 7 or 11, you list out all the pairs of numbers that add up to those sums. Each pair is considered a unique outcome. For a sum of 7, pairs like (1,6), (2,5), etc., are considered.
Knowing all the possible outcomes and identifying the combinations that lead to particular sums is the first step in calculating probabilities based on these outcomes.
Calculating Probabilities
To calculate the probability of an event, you divide the number of successful outcomes by the total number of possible outcomes. This simple ratio gives you a chance of an event occurring.
For example, calculating the probability of rolling a sum of 7 on two dice involves first identifying all possible die combinations for that sum, numbering 6 in total.
The probability then is \(\frac{6}{36} = \frac{1}{6}\), given there are 36 possible outcomes when rolling two dice. Similarly, there are 2 combinations for a sum of 11, so the probability is \(\frac{2}{36} = \frac{1}{18}\).

When dealing with mutually exclusive events, such as the sum of 7 or 11, you simply add their individual probabilities. This is possible because they cannot happen simultaneously:
  • The probability of getting a sum of 7 or 11 is calculated as \(\frac{1}{6} + \frac{1}{18}\).
  • This requires finding a common denominator to combine them into a single probability \(\frac{2}{9}\).
Thus, the calculations are streamlined using basic arithmetic rules, helping derive accurate probabilities for various outcomes.

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Most popular questions from this chapter

You draw two cards from a standard deck of 52 cards, but before you draw the second card, you put the first one back and reshuffle the deck. (a) Are the outcomes on the two cards independent? Why? (b) Find \(P(3\) on 1 st card and 10 on 2 nd ). (c) Find \(P(10\) on 1 st card and 3 on 2 nd ). (d) Find the probability of drawing a 10 and a 3 in either order.

You toss a pair of dice. (a) Determine the number of possible pairs of outcomes. (Recall that there are six possible outcomes for each die.) (b) There are three even numbers on each die. How many outcomes are possible with even numbers appearing on each die? (c) Probability extension: What is the probability that both dice will show an even number?

What is the main difference between a situation in which the use of the permutations rule is appropriate and one in which the use of the combinations rule is appropriate?

In a sales effectiveness seminar, a group of sales representatives tried two approaches to selling a customer a new automobile: the aggressive approach and the passive approach. For 1160 customers, the following record was kept: $$\begin{array}{lllr} \hline & \text { Sale } & \text { No Sale } & \text { Row Total } \\ \hline \text { Aggressive } & 270 & 310 & 580 \\ \text { Passive } & 416 & 164 & 580 \\ \text { Column Total } & 686 & 474 & 1160 \\ \hline \end{array}$$ Suppose a customer is selected at random from the 1160 participating customers. Let us use the following notation for events: \(A=\) aggressive approach, \(P a=\) passive approach, \(S=\) sale, \(N=\) no sale. So, \(P(A)\) is the probability that an aggressive approach was used, and so on. (a) Compute \(P(S), P(S \mid A)\), and \(P(S \mid P a)\). (b) Are the events \(S=\) sale and \(P a=\) passive approach independent? Explain. (c) Compute \(P(A\) and \(S)\) and \(P(P a\) and \(S)\). (d) Compute \(P(N)\) and \(P(N \mid A)\). (e) Are the events \(N=\) no sale and \(A=\) aggressive approach independent? Explain. (f) Compute \(P(A\) or \(S)\).

Sometimes probability statements are expressed in terms of odds. The odds in favor of an event \(A\) is the ratio \(\frac{P(A)}{P(n o t A)}=\frac{P(A)}{P\left(A^{c}\right)}\). For instance, if \(P(A)=0.60\), then \(P\left(A^{c}\right)=0.40\) and the odds in favor of \(A\) are \(\frac{0.60}{0.40}=\frac{6}{4}=\frac{3}{2}\), written as 3 to 2 or \(3: 2 .\) (a) Show that if we are given the odds in favor of event \(A\) as \(n: m\), the probability of event \(A\) is given by \(P(A)=\frac{n}{n+m} \cdot\) Hint \(:\) Solve the equation \(\frac{n}{m}=\frac{P(A)}{1-P(A)}\) for \(P(\bar{A})\). (b) A telemarketing supervisor tells a new worker that the odds of making a sale on a single call are 2 to \(15 .\) What is the probability of a successful call? (c) A sports announcer says that the odds a basketball player will make a free throw shot are 3 to \(5 .\) What is the probability the player will make the shot?

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