Possible Outcomes for the First Two Tosses

Oct 13, 2025 by miodeya.com 38 views

Introduction

Probability is a fundamental concept in mathematics that deals with the likelihood of an event occurring. In this article, we will explore the concept of probability in the context of coin tosses. Specifically, we will examine the probability of getting heads on only the last toss when a fair coin is tossed 3 times in a row.

What is a Fair Coin?

A fair coin is a coin that has an equal probability of landing on either heads or tails. In other words, the probability of getting heads is 1/2, and the probability of getting tails is also 1/2.

The Problem

We are given a fair coin that is tossed 3 times in a row. We want to find the probability that heads appears on only the last toss.

Breaking Down the Problem

To solve this problem, we need to consider the possible outcomes of the first two tosses. Since each toss has two possible outcomes (heads or tails), there are a total of 2 x 2 = 4 possible outcomes for the first two tosses.

Possible Outcomes for the First Two Tosses

Here are the possible outcomes for the first two tosses:

Tails, Tails (TT)
Tails, Heads (TH)
Heads, Tails (HT)
Heads, Heads (HH)

Probability of Getting Heads on the Last Toss

Now, let's consider the probability of getting heads on the last toss for each of the possible outcomes of the first two tosses.

If the first two tosses are TT, the probability of getting heads on the last toss is 1/2.
If the first two tosses are TH, the probability of getting heads on the last toss is 1/2.
If the first two tosses are HT, the probability of getting heads on the last toss is 1/2.
If the first two tosses are HH, the probability of getting heads on the last toss is 1/2.

Calculating the Total Probability

Since each of the possible outcomes of the first two tosses is equally likely, we can calculate the total probability of getting heads on the last toss by taking the average of the probabilities for each outcome.

The probability of getting heads on the last toss is therefore:

(1/2 + 1/2 + 1/2 + 1/2) / 4 = 2/4 = 1/2

However, this is not the correct answer. We need to consider the fact that we are only interested in the cases where heads appears on the last toss.

Correcting the Calculation

Let's re-examine the possible outcomes of the first two tosses:

TT: The probability of getting heads on the last toss is 1/2.
TH: The probability of getting heads on the last toss is 1/2.
HT: The probability of getting heads on the last toss is 1/2.
HH: The probability of getting heads on the last toss is 1/2.

However, we are only interested in the cases where heads appears on the last toss. Therefore, we need to consider the cases where the first two tosses are either TT or TH.

The probability of getting heads on the last toss is therefore:

(1/2 + 1/2) / 2 = 1/2

However, this is still not the correct answer. We need to consider the fact that there are three possible ways to get heads on the last toss:

The probability of getting heads on the last toss is therefore:

(1/8 + 1/8 + 1/8) = 3/8

Conclusion

In conclusion, the probability of getting heads on only the last toss when a fair coin is tossed 3 times in a row is 3/8.

Answer

The correct answer is B. $\frac{3}{8}$ .

Final Thoughts

Q: What is the probability of getting heads on the last toss when a fair coin is tossed 4 times in a row?

A: To solve this problem, we need to consider the possible outcomes of the first three tosses. Since each toss has two possible outcomes (heads or tails), there are a total of 2 x 2 x 2 = 8 possible outcomes for the first three tosses. We can then calculate the probability of getting heads on the last toss for each of these outcomes and take the average.

The probability of getting heads on the last toss is therefore:

(1/8 + 1/8 + 1/8 + 1/8 + 1/8 + 1/8 + 1/8 + 1/8) = 1/2

Q: What is the probability of getting tails on the first two tosses and heads on the last toss when a fair coin is tossed 3 times in a row?

A: To solve this problem, we need to consider the possible outcomes of the first two tosses. Since each toss has two possible outcomes (heads or tails), there are a total of 2 x 2 = 4 possible outcomes for the first two tosses. We can then calculate the probability of getting tails on the first two tosses and heads on the last toss for each of these outcomes.

The probability of getting tails on the first two tosses and heads on the last toss is therefore:

(1/4 + 1/4 + 1/4 + 1/4) = 1/2

Q: What is the probability of getting heads on the first two tosses and tails on the last toss when a fair coin is tossed 3 times in a row?

A: To solve this problem, we need to consider the possible outcomes of the first two tosses. Since each toss has two possible outcomes (heads or tails), there are a total of 2 x 2 = 4 possible outcomes for the first two tosses. We can then calculate the probability of getting heads on the first two tosses and tails on the last toss for each of these outcomes.

The probability of getting heads on the first two tosses and tails on the last toss is therefore:

(1/4 + 1/4 + 1/4 + 1/4) = 1/2

Q: What is the probability of getting heads on the first toss and tails on the last two tosses when a fair coin is tossed 3 times in a row?

A: To solve this problem, we need to consider the possible outcomes of the first toss. Since each toss has two possible outcomes (heads or tails), there are a total of 2 possible outcomes for the first toss. We can then calculate the probability of getting heads on the first toss and tails on the last two tosses for each of these outcomes.

The probability of getting heads on the first toss and tails on the last two tosses is therefore:

(1/2 + 1/2) = 1

Q: What is the probability of getting tails on the first toss and heads on the last two tosses when a fair coin is tossed 3 times in a row?

A: To solve this problem, we need to consider the possible outcomes of the first toss. Since each toss has two possible outcomes (heads or tails), there are a total of 2 possible outcomes for the first toss. We can then calculate the probability of getting tails on the first toss and heads on the last two tosses for each of these outcomes.

The probability of getting tails on the first toss and heads on the last two tosses is therefore:

(1/2 + 1/2) = 1

Q: What is the probability of getting heads on the first two tosses and tails on the last toss when a fair coin is tossed 4 times in a row?

A: To solve this problem, we need to consider the possible outcomes of the first three tosses. Since each toss has two possible outcomes (heads or tails), there are a total of 2 x 2 x 2 = 8 possible outcomes for the first three tosses. We can then calculate the probability of getting heads on the first two tosses and tails on the last toss for each of these outcomes.

The probability of getting heads on the first two tosses and tails on the last toss is therefore:

(1/8 + 1/8 + 1/8 + 1/8 + 1/8 + 1/8 + 1/8 + 1/8) = 1/2

Q: What is the probability of getting tails on the first two tosses and heads on the last two tosses when a fair coin is tossed 4 times in a row?

A: To solve this problem, we need to consider the possible outcomes of the first three tosses. Since each toss has two possible outcomes (heads or tails), there are a total of 2 x 2 x 2 = 8 possible outcomes for the first three tosses. We can then calculate the probability of getting tails on the first two tosses and heads on the last two tosses for each of these outcomes.

The probability of getting tails on the first two tosses and heads on the last two tosses is therefore:

(1/8 + 1/8 + 1/8 + 1/8 + 1/8 + 1/8 + 1/8 + 1/8) = 1/2

Conclusion

In conclusion, we have explored the concept of probability in the context of coin tosses. We have answered several frequently asked questions and provided the correct probabilities for each scenario. We hope that this article has been helpful in understanding the concept of probability and how it can be applied to real-world situations.