QuestionConsider flipping a weighted coin with outcomes $S=\{H, T\}$
Suppose that heads is 3 times as likely to occur as tails, that is $P(H)=3 P(T)$ . Which of the following statements is true? a. There is not enough information to determine the value of either $P(T)$ or $P(H)$ . b. $\quad P(T)=\frac{3}{4}$ c. $P(T)=\frac{1}{4}$ d. $P(H)=\frac{1}{4}$

Studdy Solution

STEP 1

What is this asking? We're figuring out the probability of getting heads and tails with a weighted coin, knowing heads is three times more likely than tails. Watch out! Don't mix up the probabilities of heads and tails, or forget that the total probability of all outcomes must add to one!

STEP 2

1. Set up the equation.
2. Solve for $P(T)$ .
3. Calculate $P(H)$ .
4. Check the answer.

STEP 3

Alright, so we know that heads ( $H$ ) is three times more likely than tails ( $T$ ).
We can write this as $P(H) = 3 \cdot P(T)$ .
This is our first key equation!

STEP 4

We also know that the total probability of all possible outcomes must equal $\textbf{1}$ .
Since we only have two outcomes, heads and tails, we can write this as $P(H) + P(T) = 1$ .
This is our second key equation!

STEP 5

Now, let's do some substitution magic!
We know $P(H) = 3 \cdot P(T)$ from our first equation.
Let's plug this into our second equation: $(3 \cdot P(T)) + P(T) = 1$ .

STEP 6

Combining the terms with $P(T)$ gives us $4 \cdot P(T) = 1$ .

STEP 7

To isolate $P(T)$ , we divide both sides of the equation by $\textbf{4}$ : $\frac{4 \cdot P(T)}{4} = \frac{1}{4}$ $P(T) = \frac{1}{4}$ So, the probability of getting tails is $\frac{\textbf{1}}{\textbf{4}}$ !

STEP 8

Remember our first equation, $P(H) = 3 \cdot P(T)$ ?
Now that we know $P(T) = \frac{1}{4}$ , we can easily find $P(H)$ !

STEP 9

Substitute $P(T) = \frac{1}{4}$ into the equation: $P(H) = 3 \cdot \frac{1}{4}$ .

STEP 10

Multiply to find $P(H) = \frac{\textbf{3}}{\textbf{4}}$ .
The probability of getting heads is $\frac{3}{4}$ !

STEP 11

Let's make sure everything adds up!
We found $P(H) = \frac{3}{4}$ and $P(T) = \frac{1}{4}$ .

STEP 12

Adding them together: $\frac{3}{4} + \frac{1}{4} = \frac{3+1}{4} = \frac{4}{4} = 1$ .
Perfect! The total probability is indeed $\textbf{1}$ , just as it should be.

STEP 13

The probability of tails is $P(T) = \frac{1}{4}$ and the probability of heads is $P(H) = \frac{3}{4}$ .
Therefore, the correct answer is c.

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Studdy Solution

STEP 1

What is this asking? We're figuring out the probability of getting heads and tails with a weighted coin, knowing heads is three times more likely than tails. Watch out! Don't mix up the probabilities of heads and tails, or forget that the total probability of all outcomes must add to one!

STEP 2

1. Set up the equation.
2. Solve for $P(T)$ .
3. Calculate $P(H)$ .
4. Check the answer.

STEP 3

Alright, so we know that heads ( $H$ ) is three times more likely than tails ( $T$ ).
We can write this as $P(H) = 3 \cdot P(T)$ .
This is our first key equation!

STEP 4

We also know that the total probability of all possible outcomes must equal $\textbf{1}$ .
Since we only have two outcomes, heads and tails, we can write this as $P(H) + P(T) = 1$ .
This is our second key equation!

STEP 5

Now, let's do some substitution magic!
We know $P(H) = 3 \cdot P(T)$ from our first equation.
Let's plug this into our second equation: $(3 \cdot P(T)) + P(T) = 1$ .

STEP 6

Combining the terms with $P(T)$ gives us $4 \cdot P(T) = 1$ .

STEP 7

To isolate $P(T)$ , we divide both sides of the equation by $\textbf{4}$ : $\frac{4 \cdot P(T)}{4} = \frac{1}{4}$ $P(T) = \frac{1}{4}$ So, the probability of getting tails is $\frac{\textbf{1}}{\textbf{4}}$ !

STEP 8

Remember our first equation, $P(H) = 3 \cdot P(T)$ ?
Now that we know $P(T) = \frac{1}{4}$ , we can easily find $P(H)$ !

STEP 9

Substitute $P(T) = \frac{1}{4}$ into the equation: $P(H) = 3 \cdot \frac{1}{4}$ .

STEP 10

Multiply to find $P(H) = \frac{\textbf{3}}{\textbf{4}}$ .
The probability of getting heads is $\frac{3}{4}$ !

STEP 11

Let's make sure everything adds up!
We found $P(H) = \frac{3}{4}$ and $P(T) = \frac{1}{4}$ .

STEP 12

Adding them together: $\frac{3}{4} + \frac{1}{4} = \frac{3+1}{4} = \frac{4}{4} = 1$ .
Perfect! The total probability is indeed $\textbf{1}$ , just as it should be.

STEP 13

The probability of tails is $P(T) = \frac{1}{4}$ and the probability of heads is $P(H) = \frac{3}{4}$ .
Therefore, the correct answer is c.

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Studdy Solution

STEP 1

What is this asking? We're figuring out the probability of getting heads and tails with a weighted coin, knowing heads is three times more likely than tails. Watch out! Don't mix up the probabilities of heads and tails, or forget that the total probability of all outcomes must add to one!

STEP 2

1. Set up the equation.2. Solve for P(T)P(T)P(T).3. Calculate P(H)P(H)P(H).4. Check the answer.

STEP 3

Alright, so we know that heads (HHH) is **three times** more likely than tails (TTT).We can write this as P(H)=3⋅P(T)P(H) = 3 \cdot P(T)P(H)=3⋅P(T).This is our **first key equation**!

STEP 4

We also know that the **total probability** of all possible outcomes must equal 1\textbf{1}1.Since we only have two outcomes, heads and tails, we can write this as P(H)+P(T)=1P(H) + P(T) = 1P(H)+P(T)=1.This is our **second key equation**!

STEP 5

Now, let's do some substitution magic!We know P(H)=3⋅P(T)P(H) = 3 \cdot P(T)P(H)=3⋅P(T) from our first equation.Let's plug this into our second equation: (3⋅P(T))+P(T)=1(3 \cdot P(T)) + P(T) = 1(3⋅P(T))+P(T)=1.

STEP 6

Combining the terms with P(T)P(T)P(T) gives us 4⋅P(T)=14 \cdot P(T) = 14⋅P(T)=1.

STEP 7

To isolate P(T)P(T)P(T), we **divide both sides** of the equation by 4\textbf{4}4: 4⋅P(T)4=14 \frac{4 \cdot P(T)}{4} = \frac{1}{4} 44⋅P(T)​=41​ P(T)=14 P(T) = \frac{1}{4} P(T)=41​So, the probability of getting tails is 14\frac{\textbf{1}}{\textbf{4}}41​!

STEP 8

Remember our **first equation**, P(H)=3⋅P(T)P(H) = 3 \cdot P(T)P(H)=3⋅P(T)?Now that we know P(T)=14P(T) = \frac{1}{4}P(T)=41​, we can easily find P(H)P(H)P(H)!

STEP 9

Substitute P(T)=14P(T) = \frac{1}{4}P(T)=41​ into the equation: P(H)=3⋅14P(H) = 3 \cdot \frac{1}{4}P(H)=3⋅41​.

STEP 10

Multiply to find P(H)=34P(H) = \frac{\textbf{3}}{\textbf{4}}P(H)=43​.The probability of getting heads is 34\frac{3}{4}43​!

STEP 11

Let's make sure everything adds up!We found P(H)=34P(H) = \frac{3}{4}P(H)=43​ and P(T)=14P(T) = \frac{1}{4}P(T)=41​.

STEP 12

Adding them together: 34+14=3+14=44=1\frac{3}{4} + \frac{1}{4} = \frac{3+1}{4} = \frac{4}{4} = 143​+41​=43+1​=44​=1.Perfect! The total probability is indeed 1\textbf{1}1, just as it should be.

STEP 13

The probability of tails is P(T)=14P(T) = \frac{1}{4}P(T)=41​ and the probability of heads is P(H)=34P(H) = \frac{3}{4}P(H)=43​.Therefore, the correct answer is c.

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1. Set up the equation.
2. Solve for $P(T)$ .
3. Calculate $P(H)$ .
4. Check the answer.

Alright, so we know that heads ( $H$ ) is three times more likely than tails ( $T$ ).
We can write this as $P(H) = 3 \cdot P(T)$ .
This is our first key equation!

We also know that the total probability of all possible outcomes must equal $\textbf{1}$ .
Since we only have two outcomes, heads and tails, we can write this as $P(H) + P(T) = 1$ .
This is our second key equation!

Now, let's do some substitution magic!
We know $P(H) = 3 \cdot P(T)$ from our first equation.
Let's plug this into our second equation: $(3 \cdot P(T)) + P(T) = 1$ .

Combining the terms with $P(T)$ gives us $4 \cdot P(T) = 1$ .

To isolate $P(T)$ , we divide both sides of the equation by $\textbf{4}$ : $\frac{4 \cdot P(T)}{4} = \frac{1}{4}$ $P(T) = \frac{1}{4}$ So, the probability of getting tails is $\frac{\textbf{1}}{\textbf{4}}$ !

Remember our first equation, $P(H) = 3 \cdot P(T)$ ?
Now that we know $P(T) = \frac{1}{4}$ , we can easily find $P(H)$ !

Substitute $P(T) = \frac{1}{4}$ into the equation: $P(H) = 3 \cdot \frac{1}{4}$ .

Multiply to find $P(H) = \frac{\textbf{3}}{\textbf{4}}$ .
The probability of getting heads is $\frac{3}{4}$ !

Let's make sure everything adds up!
We found $P(H) = \frac{3}{4}$ and $P(T) = \frac{1}{4}$ .

Adding them together: $\frac{3}{4} + \frac{1}{4} = \frac{3+1}{4} = \frac{4}{4} = 1$ .
Perfect! The total probability is indeed $\textbf{1}$ , just as it should be.

The probability of tails is $P(T) = \frac{1}{4}$ and the probability of heads is $P(H) = \frac{3}{4}$ .
Therefore, the correct answer is c.