Math / Data & Statistics

Question12. When four fair coins are tossed, what is the probability that the outcome will consist of two heads and two tails? (A) $\frac{1}{2}$ (B) $\frac{11}{20}$ (C) $\frac{1}{3}$ (D) $\frac{1}{4}$ (E) $\frac{3}{8}$

Studdy Solution

STEP 1

What is this asking? If we flip four coins, what are the chances we get exactly two heads and two tails? Watch out! Don't just think about one way to get two heads and two tails (like HHTT).
There are other ways to get the same result!

STEP 2

1. Find the total number of possible outcomes.
2. Find the number of ways to get two heads and two tails.
3. Calculate the probability.

STEP 3

Each coin flip has two possible outcomes: heads (H) or tails (T).
Since we're flipping four coins, we multiply the number of outcomes for each coin together.
It's like choosing from a menu!
Two options for the first coin, two for the second, two for the third, and two for the fourth!

STEP 4

So, the total number of outcomes is $2 \cdot 2 \cdot 2 \cdot 2 = 2^4 = \textbf{16}$ .
Sixteen different ways those coins can land!

STEP 5

We need to figure out how many different ways we can arrange two heads and two tails.
Think of it like picking two spots for the heads out of the four available spots.
The other two spots will automatically be tails!

STEP 6

We can use combinations for this!
The formula for combinations is $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ , where $n$ is the total number of items and $k$ is the number we're choosing.
Here, $n = 4$ (total coin flips) and $k = 2$ (number of heads we want).

STEP 7

Let's calculate: $\binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4!}{2!2!} = \frac{4 \cdot 3 \cdot 2 \cdot 1}{(2 \cdot 1)(2 \cdot 1)} = \frac{24}{4} = \textbf{6}$ So, there are six ways to get two heads and two tails.
They are HHTT, HTHT, HTTH, THHT, THTH, and TTHH.

STEP 8

Probability is the number of desired outcomes divided by the total number of possible outcomes.
We've got both those numbers now!

STEP 9

Our desired outcomes (two heads and two tails) is $\textbf{6}$ .
The total possible outcomes is $\textbf{16}$ .

STEP 10

So, the probability is $\frac{6}{16}$ .
We can simplify this fraction by dividing both the numerator and the denominator by their greatest common factor, which is 2.
This gives us $\frac{6 \div 2}{16 \div 2} = \frac{3}{8}$ .

STEP 11

The probability of getting two heads and two tails when flipping four fair coins is $\frac{3}{8}$ , which corresponds to answer choice (E).

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Question12. When four fair coins are tossed, what is the probability that the outcome will consist of two heads and two tails? (A) $\frac{1}{2}$ (B) $\frac{11}{20}$ (C) $\frac{1}{3}$ (D) $\frac{1}{4}$ (E) $\frac{3}{8}$

Studdy Solution

STEP 1

What is this asking? If we flip four coins, what are the chances we get exactly two heads and two tails? Watch out! Don't just think about one way to get two heads and two tails (like HHTT).
There are other ways to get the same result!

STEP 2

1. Find the total number of possible outcomes.
2. Find the number of ways to get two heads and two tails.
3. Calculate the probability.

STEP 3

Each coin flip has two possible outcomes: heads (H) or tails (T).
Since we're flipping four coins, we multiply the number of outcomes for each coin together.
It's like choosing from a menu!
Two options for the first coin, two for the second, two for the third, and two for the fourth!

STEP 4

So, the total number of outcomes is $2 \cdot 2 \cdot 2 \cdot 2 = 2^4 = \textbf{16}$ .
Sixteen different ways those coins can land!

STEP 5

We need to figure out how many different ways we can arrange two heads and two tails.
Think of it like picking two spots for the heads out of the four available spots.
The other two spots will automatically be tails!

STEP 6

We can use combinations for this!
The formula for combinations is $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ , where $n$ is the total number of items and $k$ is the number we're choosing.
Here, $n = 4$ (total coin flips) and $k = 2$ (number of heads we want).

STEP 7

Let's calculate: $\binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4!}{2!2!} = \frac{4 \cdot 3 \cdot 2 \cdot 1}{(2 \cdot 1)(2 \cdot 1)} = \frac{24}{4} = \textbf{6}$ So, there are six ways to get two heads and two tails.
They are HHTT, HTHT, HTTH, THHT, THTH, and TTHH.

STEP 8

Probability is the number of desired outcomes divided by the total number of possible outcomes.
We've got both those numbers now!

STEP 9

Our desired outcomes (two heads and two tails) is $\textbf{6}$ .
The total possible outcomes is $\textbf{16}$ .

STEP 10

So, the probability is $\frac{6}{16}$ .
We can simplify this fraction by dividing both the numerator and the denominator by their greatest common factor, which is 2.
This gives us $\frac{6 \div 2}{16 \div 2} = \frac{3}{8}$ .

STEP 11

The probability of getting two heads and two tails when flipping four fair coins is $\frac{3}{8}$ , which corresponds to answer choice (E).

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Question12. When four fair coins are tossed, what is the probability that the outcome will consist of two heads and two tails? (A) 12\frac{1}{2}21​ (B) 1120\frac{11}{20}2011​ (C) 13\frac{1}{3}31​ (D) 14\frac{1}{4}41​ (E) 38\frac{3}{8}83​

Studdy Solution

STEP 1

What is this asking? If we flip four coins, what are the chances we get exactly two heads and two tails? Watch out! Don't just think about *one* way to get two heads and two tails (like HHTT).There are other ways to get the same result!

STEP 2

1. Find the total number of possible outcomes.2. Find the number of ways to get two heads and two tails.3. Calculate the probability.

STEP 3

Each coin flip has two possible outcomes: heads (H) or tails (T).Since we're flipping four coins, we multiply the number of outcomes for each coin together.It's like choosing from a menu!Two options for the first coin, two for the second, two for the third, and two for the fourth!

STEP 4

So, the **total number of outcomes** is 2⋅2⋅2⋅2=24=162 \cdot 2 \cdot 2 \cdot 2 = 2^4 = \textbf{16}2⋅2⋅2⋅2=24=16.Sixteen different ways those coins can land!

STEP 5

We need to figure out how many different ways we can arrange two heads and two tails.Think of it like picking two spots for the heads out of the four available spots.The other two spots will automatically be tails!

STEP 6

We can use combinations for this!The formula for combinations is (nk)=n!k!(n−k)!\binom{n}{k} = \frac{n!}{k!(n-k)!}(kn​)=k!(n−k)!n!​, where nnn is the total number of items and kkk is the number we're choosing.Here, n=4n = 4n=4 (total coin flips) and k=2k = 2k=2 (number of heads we want).

STEP 7

STEP 8

Probability is the number of **desired outcomes** divided by the **total number of possible outcomes**.We've got both those numbers now!

STEP 9

Our **desired outcomes** (two heads and two tails) is 6\textbf{6}6.The **total possible outcomes** is 16\textbf{16}16.

STEP 10

So, the probability is 616\frac{6}{16}166​.We can simplify this fraction by dividing both the numerator and the denominator by their greatest common factor, which is 2.This gives us 6÷216÷2=38\frac{6 \div 2}{16 \div 2} = \frac{3}{8}16÷26÷2​=83​.

STEP 11

The probability of getting two heads and two tails when flipping four fair coins is 38\frac{3}{8}83​, which corresponds to answer choice (E).

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Question12. When four fair coins are tossed, what is the probability that the outcome will consist of two heads and two tails? (A) $\frac{1}{2}$ (B) $\frac{11}{20}$ (C) $\frac{1}{3}$ (D) $\frac{1}{4}$ (E) $\frac{3}{8}$

What is this asking? If we flip four coins, what are the chances we get exactly two heads and two tails? Watch out! Don't just think about one way to get two heads and two tails (like HHTT).
There are other ways to get the same result!

1. Find the total number of possible outcomes.
2. Find the number of ways to get two heads and two tails.
3. Calculate the probability.

Each coin flip has two possible outcomes: heads (H) or tails (T).
Since we're flipping four coins, we multiply the number of outcomes for each coin together.
It's like choosing from a menu!
Two options for the first coin, two for the second, two for the third, and two for the fourth!

So, the total number of outcomes is $2 \cdot 2 \cdot 2 \cdot 2 = 2^4 = \textbf{16}$ .
Sixteen different ways those coins can land!

We need to figure out how many different ways we can arrange two heads and two tails.
Think of it like picking two spots for the heads out of the four available spots.
The other two spots will automatically be tails!

We can use combinations for this!
The formula for combinations is $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ , where $n$ is the total number of items and $k$ is the number we're choosing.
Here, $n = 4$ (total coin flips) and $k = 2$ (number of heads we want).

Probability is the number of desired outcomes divided by the total number of possible outcomes.
We've got both those numbers now!

Our desired outcomes (two heads and two tails) is $\textbf{6}$ .
The total possible outcomes is $\textbf{16}$ .

So, the probability is $\frac{6}{16}$ .
We can simplify this fraction by dividing both the numerator and the denominator by their greatest common factor, which is 2.
This gives us $\frac{6 \div 2}{16 \div 2} = \frac{3}{8}$ .

The probability of getting two heads and two tails when flipping four fair coins is $\frac{3}{8}$ , which corresponds to answer choice (E).