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Given that we have already tossed a balanced coin ten times and obtained zero heads, what is the probability that we must toss it at least two more times to obtain the first head?

Short Answer

Expert verified
The probability is 0.25.

Step by step solution

01

Understanding the Problem

We need to find the probability that after obtaining zero heads in ten coin tosses, at least two more tosses are needed to get the first head.
02

Define the Events

Let's define the event A as getting the first head in one toss, and event B as not getting a head in the first toss after 10 failures. Thus, the probability of event A is the probability of getting one head in one toss, which is 0.5, and event B is the probability of obtaining tails, which is also 0.5.
03

Calculate the Probability of Two Tail Succession

Since we need the first two additional tosses to not yield a head, we calculate the probability of getting tails in both of these tosses. This is given by the probability of two successive tails: \( P(T \text{ and } T) = 0.5 \times 0.5 = 0.25 \).
04

Find Probability of At Least Two More Tosses

For the first head to occur at least in the third toss, the first two should be tails. Therefore, the probability that it takes at least two more tosses to get a head is simply the probability of getting tails in both the next two tosses, which is 0.25.

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Key Concepts

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

Coin Toss Probability
Coin tosses are a classic example in probability theory due to their straightforward nature and predictable outcomes. When you toss a coin, there are two possible outcomes: Heads or Tails. Each outcome is equally likely, which means the probability of getting either heads or tails is
  • 0.5 (or 50%).
This is because the coin is assumed to be fair or balanced, meaning there is no bias towards either heads or tails.

The independence of events is key in understanding this concept. For each toss of the coin, the probability remains constant at 0.5, regardless of previous results. This means if you toss a coin ten times, as in our exercise, and you've received tails every time, the probability of the next toss being a head is still 0.5.

Understanding this base probability helps in solving more complex problems, where sequences of these outcomes play a role, such as determining the probability when expecting a certain number of heads after several tosses.
Geometric Distribution
The geometric distribution is a probability distribution that models the number of trials required for the first success in a series of Bernoulli trials. In a Bernoulli trial, each trial has exactly two possible outcomes: "success" and "failure". Applying this to our exercise, tossing a coin until you get heads is a scenario captured by the geometric distribution.
  • Here, each coin toss can be considered a Bernoulli trial.
  • The "success" is obtaining a head on a toss.
  • The probability of success (getting a head) on each toss is 0.5.
In our given scenario, where no heads have come up in ten tosses, we are interested in finding out the likelihood of needing at least two more tosses to get a head. The geometric distribution informs us that
  • The probability of getting a head for the first time on the nth toss is \(P(X = n) = (1 - p)^{n - 1} \times p\),
where \( p = 0.5 \), the probability of heads. The distribution is memoryless, so even after ten tails, each toss behaves as if starting fresh.
Independent Events
Independence is a fundamental concept in probability that refers to events that do not influence each other’s outcomes. In the context of coin tosses, each toss is independent, meaning the outcome of one toss does not affect the outcome of another.
  • For example, tossing tails ten times does not change the probability of getting heads (0.5) on any subsequent toss.
  • This independence is crucial when dealing with sequences of trials, as seen with the geometric distribution.
In our exercise, understanding that each toss is an independent event allows us to calculate compounded probabilities straightforwardly.

When calculating the probability that the first head occurs at or after the third toss, we consider the first two as tails with a probability of 0.5 each.

Because the outcome of each is independent, the probability of successive tails is the product of their probabilities. In the instance described in the solution steps,
  • The probability that both the first and second tosses after ten tails are also tails equals \( 0.5 \times 0.5 = 0.25 \).
This illustrates the core idea of independent events and their significance in probability theory.

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Most popular questions from this chapter

A store owner has overstocked a certain item and decides to use the following promotion to decrease the supply. The item has a marked price of \(100 .\) For each customer purchasing the item during a particular day, the owner will reduce the price by a factor of one-half. Thus, the first customer will pay \(50\) for the item, the second will pay \(25,\) and so on. Suppose that the number of customers who purchase the item during the day has a Poisson distribution with mean 2 . Find the expected cost of the item at the end of the day. [Hint: The cost at the end of the day is \(100(1 / 2)^{Y}\), where \(Y\) is the number of customers who have purchased the item.]

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