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Chapter 7: Problem 7

What is the probability that when a coin is flipped six times in a row, it lands heads up every time?

Short Answer

Expert verified
The probability is \(\frac{1}{64}\).

Step by step solution

01

Understand the Problem

Determine the number of coin flips and what outcome is desired. In this case, a coin is flipped 6 times, and the desired outcome is getting heads on each flip.
02

Probability of a Single Flip

Calculate the probability of getting heads in a single coin flip. Since a fair coin has two sides, the probability of landing heads in one flip is \(\frac{1}{2}\).
03

Probability of Consecutive Events

For independent events, multiply the probability of each event. Since each coin flip is independent, multiply \(\frac{1}{2}\) for each of the 6 flips: \(\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}\).
04

Simplify the Expression

Simplify the expression: \(\frac{1}{2^6} = \frac{1}{64}\). This is the probability of getting heads 6 times in a row.

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Key Concepts

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

Independent Events
In probability theory, independent events are events where the occurrence of one event does not affect the probability of the other. This means that each event happens separately. In our example, each coin flip is an independent event.
When a coin is flipped, the outcome of one flip (heads or tails) does not influence the outcome of subsequent flips.
For example:
  • If you flip a coin and get heads, the probability of getting heads or tails on the next flip is still 50%.
  • This is true for each flip in a series of flips, such as flipping the coin 6 times as in our problem.
Understanding the concept of independent events helps us to calculate the combined probability of multiple events (like consecutive coin flips) happening.
Probability of Consecutive Events
Calculating the probability of multiple independent events happening in a sequence involves multiplying the probability of each individual event.
Let's use the coin flipping example: each flip of the coin has a probability of \(\frac{1}{2}\) for landing heads.
If we want the coin to land heads 6 times in a row, we multiply the probability of each individual event:
  • On the first flip: \(\frac{1}{2}\)
  • On the second flip: \(\frac{1}{2}\)
This continues for each of the 6 flips:
  • \( \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \)
  • \( \frac{1}{2^6} \) which simplifies to \( \frac{1}{64} \).
Thus, the probability of getting heads 6 times in a row is \( \frac{1}{64} \).
Simplifying Fractions
Simplifying fractions is a crucial step in many probability calculations.
To simplify a fraction, you divide the numerator (top number) and the denominator (bottom number) by any common factors they have.
In our coin-flipping example, the fraction comes from calculating the probability of 6 independent events (each with probability \( \frac{1}{2}\)) occurring consecutively:
  • We started with \( \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \).
  • This multiplication simplifies to \( \frac{1}{2^6} \).
  • Calculating \( 2^6 \) gives us 64, so \( \frac{1}{2^6} \) simplifies to \( \frac{1}{64} \).
Thus, the probability of flipping 6 heads in a row is \( \frac{1}{64} \). Simplifying fractions makes it easier to understand and communicate the probability results.

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Most popular questions from this chapter

A player in the Powerball lottery picks five different integers between 1 and 69 , inclusive, and a sixth integer between 1 and \(26,\) which may duplicate one of the earlier five integers. The player wins the jackpot if all six numbers match the numbers drawn. a) What is the probability that a player wins the jackpot? b) What is the probability that a player wins \(\$ 1,000,000\) , which is the prize for matching the first five numbers, but not the sixth number, drawn? c) What is the probability that a player wins \(\$ 100\) by matching exactly three of the first five and the sixth numbers drawn, or four of the first five numbers, but not the sixth number, drawn? d) What is the probability that a player wins a prize of \(\$ 4,\) which is the prize when the player matches the sixth number, and either one or none of the first five numbers drawn?

What is the conditional probability that exactly four heads appear when a fair coin is flipped five times, given that the first flip came up heads?

Which is more likely: rolling a total of 9 when two dice are rolled or rolling a total of 9 when three dice are rolled?

A pair of dice is rolled in a remote location and when you ask an honest observer whether at least one die came up six, this honest observer answers in the affirmative. a) What is the probability that the sum of the numbers that came up on the two dice is seven, given the information provided by the honest observer? b) Suppose that the honest observer tells us that at least one die came up five. What is the probability the sum of the numbers that came up on the dice is seven, given this information?

What is the expected value when a \(\$ 1\) lottery ticket is bought in which the purchaser wins exactly \(\$ 10\) million if the ticket contains the six winning numbers chosen from the set \(\\{1,2,3, \ldots, 50\\}\) and the purchaser wins nothing otherwise?
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