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Exploring combinations. A coin is tossed 5 times. How many sequences / combinations of Heads/Tails are there that have: (a) Exactly 1 Tail? (b) Exactly 4 Tails? (c) Eactly 3 Tails? (d) At least 3 Tails?

Short Answer

Expert verified
(a) 5; (b) 5; (c) 10; (d) 16.

Step by step solution

01

Understanding the Setup

We have a coin that is tossed 5 times, resulting in sequences of outcomes where each toss can be a Head (H) or a Tail (T). We need to find combinations or sequences that fit certain criteria for each part of the question.
02

Exactly 1 Tail

To have exactly 1 Tail in the sequence of 5 coin tosses means that there is 1 T and the remaining 4 positions are filled with H. We choose 1 position for the Tail from the 5 available positions. The number of ways to choose 1 position from 5 is given by the combination formula, \( \binom{n}{k} \). Here, \( n = 5 \) and \( k = 1 \), so the number of sequences is \( \binom{5}{1} = 5 \).
03

Exactly 4 Tails

To have exactly 4 Tails means there's 1 Head in the sequence of 5 tosses. We choose 1 position for the Head from the 5 available positions. Using the combination formula, \( \binom{5}{1} = 5 \). Thus, there are 5 sequences with exactly 4 Tails.
04

Exactly 3 Tails

To have exactly 3 Tails, we need to choose 3 positions for the Tails from the 5 available positions. Using the combination formula, \( \binom{5}{3} \) calculates the number of ways to select these positions: \( \binom{5}{3} = 10 \). Therefore, there are 10 sequences with exactly 3 Tails.
05

At Least 3 Tails

To find sequences with at least 3 Tails, we consider 3, 4, and 5 Tails. We calculate each case and sum them: \( \binom{5}{3} = 10 \) for 3 Tails, \( \binom{5}{4} = 5 \) for 4 Tails, and \( \binom{5}{5} = 1 \) for 5 Tails. Add these values together to get the total: 10 + 5 + 1 = 16.

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Key Concepts

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

Probability
Probability helps us understand the chance of a particular outcome occurring. In coin toss experiments, each toss has two possible results: Heads or Tails. This makes it simple, as each outcome has a probability of 50%, or 0.5, assuming the coin is fair.
For multiple coin tosses, the probability of a specific sequence of outcomes can be calculated by multiplying the probability of each individual event. If we want to find the probability of getting, say, exactly 1 Tail in 5 tosses, we must first determine the number of sequences that fit this condition. Then, we multiply that by the probability of each sequence. That's where binomial coefficients and combinations come in handy to count possible sequences.
So, in context:
  • Exactly 1 Tail from 5 tosses translates into a specific arrangement of one Tail and four Heads.
  • Our task becomes determining the number of these arrangements using probability principles.
Binomial Coefficients
Binomial coefficients play a key role in combinatorics by helping us count how many ways we can select a subset from a larger set.
The formula for binomial coefficients is given by \( \binom{n}{k} \), where \( n \) is the total number of items to choose from, and \( k \) is the number of items to choose. It's calculated as: \[\binom{n}{k} = \frac{n!}{k!(n-k)!}\]
In the context of our coin toss problem:
  • To determine how many ways there are to get exactly 1 Tail in 5 tosses, we set up the formula as \( \binom{5}{1} \).
  • This tells us there are 5 arrangements or sequences possible, each comprising 1 Tail and 4 Heads.

Knowing how to use binomial coefficients can unravel many complex combinatorial problems with ease.
Coin Toss Experiments
Coin toss experiments are classic examples in probability and combinatorics due to their straightforward nature and equally likely outcomes.
Each toss of a coin is independent, meaning the result of one toss doesn't affect the others. With multiple tosses, like five in this exercise, we're exploring sequences of Heads (H) and Tails (T).
For different scenarios, like having exactly 1 Tail, exactly 4 Tails, or at least 3 Tails, we explore different sequences. The focus is often on:
  • Counting specific patterns of results.
  • Using combination calculations for how these outcomes align with our conditions.

When analyzing such experiments:
  • We leverage binomial coefficients to simplify the task of counting and thereby understanding the distribution of sequences that satisfy our conditions.
  • This exercise helps illustrate the power of combinatorial approaches in real-world probability scenarios.

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