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If we toss an honest coin 10 times, what is the probability of (a) getting 5 heads and 5 tails? (b) getting 3 heads and 7 tails?

Short Answer

Expert verified
The probability of getting 5 heads and 5 tails when tossing an honest coin 10 times is \(\frac{63}{256}\), and the probability of getting 3 heads and 7 tails is \(\frac{15}{128}\).

Step by step solution

01

Determine the number of possible outcomes

Firstly, determine the total number of possible outcomes for tossing the coin 10 times. This can be calculated as \(2^{10}\), because each toss has 2 possible outcomes - a head (H) or a tail (T) - and there are 10 independent tosses.
02

Calculate the probability of getting 5 heads and 5 tails

To get the number of ways of getting exactly 5 heads in 10 tosses, calculate the binomial coefficient \(\binom{n}{k}\), where n is the total number of tosses and k is the desired number of successful outcomes. In this case, n=10 and k=5. Using the formula for the binomial coefficient, this gives \(\binom{10}{5}=252\). The probability is then given by the number of successful outcomes divided by the total number of possible outcomes, or \(\frac{252}{2^{10}}\). Simplifying this gives the probability as \(\frac{63}{256}\).
03

Calculate the probability of getting 3 heads and 7 tails

Following a similar method to Step 2, calculate the binomial coefficient \(\binom{10}{3}\) for the number of ways of getting exactly 3 heads in 10 tosses. This gives \(\binom{10}{3}=120\). The probability is then \(\frac{120}{2^{10}}\), which simplifies to \(\frac{15}{128}\).

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Key Concepts

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

Binomial Coefficient
The binomial coefficient is a key concept in probability theory, particularly when dealing with problems that involve counting combinations. It tells us the number of ways to choose a subset of items from a larger set. Imagine you want to select a basket of fruits from a larger pile, and this coefficient helps count the different combinations you can choose.
In the context of probability, we often represent the binomial coefficient as \( \binom{n}{k} \). This notation means "n choose k," which refers to the number of ways to choose \( k \) successes from \( n \) trials. The formula used to calculate the binomial coefficient is:
\[\binom{n}{k} = \frac{n!}{k!(n-k)!}\]where \(!\) denotes a factorial, meaning the product of all positive integers up to that number.
In our exercise, we used binomial coefficients to determine the number of ways to get a specific number of heads in a series of coin tosses. For example, to find the probability of getting 5 heads in 10 tosses, we calculated \( \binom{10}{5} = 252 \). This calculation helped us understand how many different sequences can result in 5 heads and 5 tails.
Coin Toss Experiment
The coin toss experiment is a classic example in probability due to its simplicity and binary outcome. Each toss of the coin has two possible outcomes:
  • Head (H)
  • Tail (T)
This means that each toss is an independent event and does not affect the outcome of other tosses.
When you toss a coin multiple times, the number of possible outcomes increases exponentially. If you toss a coin 10 times, the total number of possible outcomes is \( 2^{10} \) because each individual toss has 2 outcomes.
In our exercise, we have repeated coin tosses which means the number of outcomes can be determined by exponentiating the number of tosses \((n)\). For tossing a coin 10 times, that results in 1,024 different possible sequences.
This simple experiment provides a basis for understanding how probability distributions can be formed, as we calculate specific outcomes like getting a certain number of heads in 10 tosses.
Probability Calculation
Probability calculation is about finding how likely a particular event is to occur. The formula to calculate the probability of an event is:
\[P(\text{event}) = \frac{ \text{Number of Successful Outcomes} }{ \text{Total Number of Possible Outcomes} }\]Let's use this formula in our coin toss example. We found the number of ways to get certain results using the binomial coefficient. In part (a) of the exercise, to calculate the probability of getting 5 heads in 10 tosses, we divided the number of those successful sequences \( \binom{10}{5} = 252 \) by the total number of possible sequences \( 2^{10} = 1024 \).
This gives us:\[P(\text{5 heads}) = \frac{252}{1024} = \frac{63}{256}\]Likewise, to find the probability of exactly 3 heads and 7 tails, the calculation goes:\[P(\text{3 heads}) = \frac{\binom{10}{3}}{ 1024 } = \frac{120}{1024} = \frac{15}{128}\]These calculations demonstrate how you can assess possible outcomes and their likelihood in any random experiment by following simple steps and using the binomial theorem.

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Most popular questions from this chapter

The Brute Force Bandits is a punk rock band planning their next concert tour. The band has a total of 30 new songs in their repertoire. (a) How many different set lists of 18 songs can the band select to play on the tour? (Hint: Assume that the order in which the songs are listed is irrelevant.) (b) How many different ways are there for the band to record a CD consisting of 18 songs chosen from the 30 new songs? (Hint: In recording a CD, the order in which the songs appear on the \(\mathrm{CD}\) is relevant.)

Find the odds of each of the following events. (a) an event \(E\) with \(\operatorname{Pr}(E)=3 / 11\) (b) an event \(E\) with \(\operatorname{Pr}(E)=0.375\)

Five candidates \((A, B, C, D,\) and \(E)\) have a chance to be selected to be on American Idol. Any subset of them (including none of them or all of them ) can be selected, and assume that the selection process is completely random (the subsets of candidates are all equally likely). Find the probability of each of the following events. (Hint: Do Exercise 16 first.) (a) \(E_{1}:{ }^{*}\) two candidates get selected." (b) \(E_{2}:\) "three candidates get selected." (c) \(E_{3}:{ }^{4}\) three candidates get selected, and \(A\) is not one of them."

There are 347 NCAA Division I college basketball teams. (a) How many different top-25 rankings are possible? [Assume that every team has a chance to be a top-25 team. (b) How many ways are there to choose 64 teams (unseeded) to make it to the NCAA tournament? [Assume every combination of 64 teams is possible.]

Suppose that you roll a pair of honest dice. If you roll a total of \(7,\) you win \(\$ 18\); if you roll a total of 11 , you win \(\$ 54\); if you roll any other total, you lose \(\$ 9 .\) Find the expected payoff for this game.

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