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A coin is tossed nine times. What is the total number of possible outcomes of the nine-toss experiment? How many elements are in the subset" 6 heads and 3 tails"? What is the probability of getting exactly 6 heads and 3 tails in nine tosses of thus unbiased coin?

Short Answer

Expert verified
The total number of possible outcomes for the nine-toss experiment is \(2^9\). The number of elements in the subset of 6 heads and 3 tails is \(\binom{9}{6}\). The probability of getting exactly 6 heads and 3 tails in nine tosses is \(\frac{\binom{9}{6}}{2^9}\).

Step by step solution

01

Determine the total number of possible outcomes

Using the counting principle, there are two possible outcomes for each individual toss (heads or tails). Since there are nine tosses, we can determine the total number of outcomes by using the following equation: \(Total\ Outcomes = 2^{No.\ of\ Tosses}\) In this case, the total number of outcomes can be represented as: \( Total\ Outcomes = 2^9 \)
02

Calculate the number of elements in the subset of 6 heads and 3 tails

To find the number of elements in the subset of exactly 6 heads and 3 tails, we can use the binomial coefficient formula. The formula helps us determine the number of ways to choose 6 heads from a total of 9 coin tosses: \[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \] where \(n\) represents the total number of tosses and \(k\) represents the number of desired heads. In our case, \(n = 9\) and \(k = 6\). Calculating the number of elements in the subset: \[ \binom{9}{6} = \frac{9!}{6!(9-6)!} \]
03

Calculate the Probability of getting exactly 6 heads and 3 tails

Now that we know the total number of outcomes and the number of desired outcomes, we can find the probability of getting exactly 6 heads and 3 tails in nine tosses using the following formula: \[ Probability = \frac{Desired\ Outcomes}{Total\ Outcomes} \] We already know from step 1 that the total number of outcomes is \( 2^9 \) and from step 2 the number of desired outcomes can be represented as \( \binom{9}{6} \). Therefore, the probability can be calculated as: \[ Probability = \frac{\binom{9}{6}}{2^9} \] Now, we just have to plug in the values and calculate the probability.

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Key Concepts

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

Binomial Coefficient
Understanding the binomial coefficient is central to solving probability problems where there are two potential outcomes for each event. In the context of a coin toss, these outcomes are typically heads or tails. But what if we’re looking at multiple coin tosses and want to find out the probability of getting a specific combination of heads and tails?

The binomial coefficient tells us the number of ways we can achieve a certain number of successes (for instance, flipping heads) in a series of independent events (such as coin tosses). It’s represented using the notation \(\binom{n}{k}\), where \(n\) is the total number of events and \(k\) is the number of successes we are interested in. It is calculated using the formula: \[\binom{n}{k} = \frac{n!}{k!(n-k)!}\] Here, the \( ! \) symbol represents factorial notation, which is another important concept in probability calculations.

For nine coin tosses, if we wish to calculate how many ways we can get exactly six heads, we plug in \(n = 9\) and \(k = 6\) into the binomial coefficient formula. This calculation provides insight into one aspect of our overall probability question.
Counting Principle
When we look at probability, the counting principle helps us understand the basic foundational rule for calculating the total number of possible outcomes for a series of events. Applied to coin tosses, the counting principle states that if one event can occur in \(m\) ways and another independent event can occur in \(n\) ways, then the total number of ways both events can occur is \(m \times n\).

If a coin has two possible outcomes for each toss and we are tossing it nine times, the counting principle suggests that each toss multiplies the number of possible outcomes by 2. Therefore, for the nine-toss experiment, we calculate the total outcomes by raising 2 to the power of 9, which is a straightforward application of the counting principle: \(Total\ Outcomes = 2^9\). This foundation is crucial for understanding the sample space, which is the set of all possible outcomes in a probability experiment.
Factorial Notation
Factorial notation is another fundamental part of understanding probability, and it recurs frequently when working with permutations, combinations, and, as we've seen, the binomial coefficient. In mathematics, the factorial of a non-negative integer \(n\), denoted by \(n!\), is the product of all positive integers less than or equal to \(n\).

For example, \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\). Factorial notation is indispensable when we calculate the number of possible ways to arrange items or, as in our coin toss scenario, the number of ways to obtain a certain number of heads out of a number of flips. In the calculation of the binomial coefficient \(\binom{9}{6}\), we use factorials to determine the number of elements in the subset of getting 6 heads out of 9 tosses.
Sample Space in Probability
The sample space in probability is a term used to define the set of all possible outcomes of a probabilistic experiment. For our coin toss example, the sample space consists of all the combinations of heads (H) and tails (T) that could occur over the nine tosses.

Imagine writing down every possible sequence of Hs and Ts for the nine coin tosses; this would give you the entire sample space for this experiment. Importantly, each outcome is unique and has an equal chance of occurring when using an unbiased coin. Calculating the size of the sample space gives us a baseline to determine probabilities for specific outcomes within that space, such as getting exactly 6 heads and 3 tails in nine tosses, which is just one subset of the entire sample space.

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Most popular questions from this chapter

A bag contains 1 white ball and 2 red balls. A ball is drawn at random. If the ball is white then it is put back in the bag along with another white ball. If the ball is red then it is put back in the bag with two extra red balls. Find the probability that the second ball drawn is red. If the second ball drawn is red, what is the probability that the first ball drawn was red?

Your company uses a pre-employment test to screen applicants for the job of repairman. The test is passed by \(60 \%\) of the applicants. Among those who pass the test \(80 \%\) complete training successfully. In an experiment, a random sample of applicants who do not pass the test is also employed. Training is successfully completed by only \(50 \%\) of this group. If no pre- employment test is used, what percentage of applicants would you expect to complete training successfully?

Find the probability that a face card is drawn on the first draw and an ace on the second in two consecutive draws, without replacement, from a standard deck of cards.

Find the probability that Event \(\mathrm{A}\), drawing a spade on a single draw from a deck of cards, and Event \(B\), rolling a total of 7 on a single roll of a pair of dice, will both occur.

An electronic device contains two easily removed subassemblies, \(\mathrm{A}\) and \(\mathrm{B}\). If the device fails, the probability that it will be necessary to replace A is 0.50. Some failures of A will damage B. If A must be replaced, the probability that B will also have to be replaced is \(0.70 .\) If it is not necessary to replace A, the probability that \(\mathrm{B}\). will have to be replaced is only \(0.10\). What percentage of all failures will you require to replace both \(\mathrm{A}\) and \(\mathrm{B}\) ?

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