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Chapter 15: Problem 2

Three coins are tossed; what is the probability that two are heads and one tails? That the first two are heads and the third tails? If at least two are heads, what is the probability that all are heads?

Short Answer

Expert verified
The probabilities are \( \frac{3}{8} \) for two heads and one tail, \( \frac{1}{8} \) for the first two heads and the third tails, and \( \frac{1}{4} \) that all are heads given at least two are heads.

Step by step solution

01

- Determine the total number of outcomes

When three coins are tossed, each coin has two possible outcomes: heads (H) or tails (T). Therefore, the total number of possible outcomes is calculated as follows: Total Outcomes = 2^3 = 8. These outcomes are: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT.
02

- Calculate the probability of getting two heads and one tail (any order)

Identify all outcomes where there are exactly two heads and one tail: - HHT - HTH - THH There are 3 such outcomes out of the total 8. The probability of getting two heads and one tail is calculated as follows: Probability = \( \frac{3}{8} \).
03

- Calculate the probability of getting heads on the first two coins and tails on the third

Count the specific outcome where the first two coins are heads and the third coin is tails: - HHT There is only 1 such outcome out of the total 8. The probability of this specific sequence happening is: Probability = \( \frac{1}{8} \).
04

- Calculate the conditional probability that all coins are heads given at least two are heads

First, identify all outcomes where at least two coins are heads: - HHH - HHT - HTH - THH There are 4 such outcomes. Out of these, only one outcome is all heads: - HHH So, the conditional probability of all coins being heads given that at least two are heads is: Conditional Probability = \( \frac{1}{4} \).

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Key Concepts

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

Coin Toss Probability
When dealing with coin tosses, each coin has two possible outcomes: heads (H) or tails (T). It's essential to understand that each outcome is equally likely. This is fundamental to calculating probabilities in coin toss scenarios.

For instance, if you toss three coins, you can think of it as a sequence of three events, where each event has an outcome of H or T. To find all possible outcomes when tossing three coins, you multiply the number of outcomes for each coin: \[ Total \, Outcomes = 2^3 = 8 \] Breaking it down visually, the outcomes are: HHH, HHT, HTH, HTT, THH, THT, TTH, and TTT.
Conditional Probability
Conditional probability is the likelihood of an event occurring given that another event has already occurred. This is particularly useful when we're considering outcomes based on certain conditions.

In our exercise, we need to find the probability that all three coins show heads (HHH) given that at least two of them are heads. First, identify all outcomes where at least two coins are heads: HHH, HHT, HTH, THH.

Out of these 4 outcomes, only HHH has all heads. Thus, the conditional probability is calculated as follows: \[ Conditional \, Probability = \frac{1}{4} \]
Outcome Calculation
Calculating outcomes accurately is essential for determining probability. In our case, we need to count specific outcomes to solve each part of the problem.

First, we list all possible outcomes of tossing three coins. These are: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT. For finding the probability of getting two heads and one tail, identify all relevant sequences, such as HHT, HTH, THH. There are 3 such sequences.

For a specific sequence like heads on the first two coins and tails on the third (HHT), note that there's only 1 such outcome among the 8 possible.
Probability Theory
Probability theory helps us quantify the likelihood of different outcomes. Understanding basic principles is crucial for solving more complex problems.

The formula for probability is given by: \[ Probability = \frac{Number \, of \, Favorable \, Outcomes}{Total \, Number \, of \, Possible \, Outcomes} \] In our exercise, we used this principle to find probabilities for different events. For example, the probability of getting two heads and one tail is \frac{3}{8}, where 3 is the number of favorable outcomes (HHT, HTH, THH) and 8 is the total number.

Mastering these basics will make tackling more complicated probability problems easier!

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Most popular questions from this chapter

(a) A weighted coin has probability \(\frac{2}{3}\) of coming up heads and probability \(\frac{1}{3}\) of coming up tails. The coin is tossed twice. Let \(x=\) number of heads. Set up the sample space for \(x\) and the associated probabilities. (b) Find \(\bar{x}\) and \(\sigma\) (c) If in (a) you know that there was at least one tail, what is the probability that both were tails?

Set up sample spaces for Problems 1 to 7 and list next to each sample point the value of the indicated random variable \(x,\) and the probability associated with the sample point. Make a table of the different values \(x_{i}\) of \(x\) and the corresponding probabilities \(p_{i}=f\left(x_{i}\right)\) Compute the mean, the variance, and the standard deviation for \(x\). Find and plot the cumulative distribution function \(F(x)\). Three coins are tossed; \(x=\) number of heads minus number of tails.

(a) A weighted coin has probability of \(\frac{2}{3}\) of showing heads and \(\frac{1}{3}\) of showing tails. Find the probabilities of \(h h, h t, t h\) and \(t t\) in two tosses of the coin. Set up the sample space and the associated probabilities. Do the probabilities add to 1 as they should? What is the probability of at least one head? What is the probability of two heads if you know there was at least one head? (b) For the coin in (a), set up the sample space for three tosses, find the associated probabilities, and use it to answer the questions in Problem 2.12 .

A bit (meaning binary digit) is 0 or \(1 .\) An ordered array of eight bits (such as 01101001) is a byte. How many different bytes are there? If you select a byte at random, what is the probability that you select \(11000010 ?\) What is the probability that you select a byte containing three 1 's and five 0 's?

Set up sample spaces for Problems 1 to 7 and list next to each sample point the value of the indicated random variable \(x,\) and the probability associated with the sample point. Make a table of the different values \(x_{i}\) of \(x\) and the corresponding probabilities \(p_{i}=f\left(x_{i}\right)\) Compute the mean, the variance, and the standard deviation for \(x\). Find and plot the cumulative distribution function \(F(x)\). A random variable \(x\) takes the values \(0,1,2,3,\) with probabilities \(\frac{5}{12}, \frac{1}{3}, \frac{1}{12}, \frac{1}{6}\).
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