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Chapter 10: Problem 13

Four fair coins are tossed. Use a multiplication tree to find the probability of the event. Exactly one of the flips is heads.

Short Answer

Expert verified
The probability is 0.25.

Step by step solution

01

Understand the Problem

Determine the total number of possible outcomes when four fair coins are tossed. Each coin has 2 outcomes: heads (H) or tails (T). The total number of outcomes is calculated as: \[ 2^4 = 16 \]
02

Identify the Favorable Outcomes

List the outcomes where exactly one of the four coin flips is heads: HTTT, THTT, TTHT, TTTH. There are 4 favorable outcomes.
03

Calculate the Probability

Use the probability formula: \[ P(\text{event}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} \] Substituting the values: \[ P(\text{exactly one head}) = \frac{4}{16} = \frac{1}{4} = 0.25 \]

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Key Concepts

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

multiplication tree
A multiplication tree helps break down a problem into simpler steps. Think of it as a branching diagram showing all possible outcomes of a sequence of events.
For our problem, we toss four fair coins. Each coin can land heads (H) or tails (T).
So, for the first coin, there are 2 outcomes: H or T.
The second coin also has 2 outcomes: H or T.
This pattern continues for the third and fourth coins.
Therefore, we have a total of \[2 \times 2 \times 2 \times 2 = 16\] outcomes.
The multiplication tree visually shows these branches and their paths. Each path represents one unique sequence of coin flips.
Understanding this method helps simplify and visualize complex probability problems.
favorable outcomes
In probability, favorable outcomes are the specific outcomes you are interested in.
For our coin toss problem, we want the outcome where exactly one of the four coin flips is heads.
List and count these favorable outcomes:
  • Heads in the first flip: HTTT
  • Heads in the second flip: THTT
  • Heads in the third flip: TTHT
  • Heads in the fourth flip: TTTH
There are exactly four favorable outcomes.
Knowing these helps us calculate the probability of our desired event occurring.
probability formula
To find the probability of an event, we use the probability formula: \[P(\text{event}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}\]
In our coin toss problem:
  • Total number of outcomes: 16 (found using the multiplication tree)
  • Number of favorable outcomes: 4 (where only one coin shows heads)
Substitute these values into the formula:
\[P(\text{exactly one head}) = \frac{4}{16} = \frac{1}{4} = 0.25\]
Therefore, the probability of flipping exactly one head in four coin tosses is 0.25 or 25%.

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Most popular questions from this chapter

Find \(k\) such that the function is a probability density function over the given interval. Then write the probability density function. If there is no \(k\) that makes the function a probability density function, state why. $$ f(x)=k x,[-1,4] $$

Let \(X\) be a continuous random variable. The \(p\) th percentile of its distribution is defined to be the number \(t\) so that the probability that \(x\) is less than or equal to \(\ell\) is \(p \%\) $$ P(X \leq t)=\frac{p}{100} $$ For example, if \(X\) has the standard normal density, then $$ P(X \leq 1.28) \approx 90 \% $$ using Table 2 . Therefore, we say that \(1.28\) is the 90 th percentile for a standard normal distribution. SAT verbal test scores are normally distributed with mean 507 and standard deviation \(111 .\) Find the following percentiles for SAT scores. \({ }^{35}\) a) 35 th percentile b) 60 th percentile c) 92 nd percentile

Find \(k\) such that the function is a probability density function over the given interval. Then write the probability density function. If there is no \(k\) that makes the function a probability density function, state why. $$ f(x)=k\left(x^{2}-x\right),[0,2] $$

Assume that the random variable \(X\) is normally distributed. Use the given information to find the unknown parameter or parameters of the distribution. If \(E(X)=-3\) and \(P(-6 \leq X \leq 0)=0.3108\), find \(\operatorname{Var}(X)\)

Verify the two properties of a probability density function over the given interval. $$ f(x)=\frac{1}{3},[4,7] $$
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