QuestionSuppose $A$ and $B$ are independent events. If $P(A)=0.4$ and $P(B)=0.9$ , what is $P\left(A^{\prime} \cap B\right)$ ? A. 0.04 B. 0.36 C. 0.06 D. 0.54

Studdy Solution

STEP 1

What is this asking? If two events are independent, and we know how likely each one is to happen, how likely is it that the first one doesn't happen and the second one does? Watch out! Don't forget what "independent events" means, and don't mix up intersection and union!

STEP 2

1. Define Independence
2. Calculate the Probability of A'
3. Calculate the Probability of A' intersect B
4. Find the Answer

STEP 3

Alright, so, independent events mean that one event happening doesn't change the chance of the other event happening.
Think of flipping a coin and rolling a die – they don't affect each other!

STEP 4

$A'$ is the complement of $A$ , meaning it's everything except $A$ .
Since the total probability is always 1, the probability of $A'$ is $1$ minus the probability of $A$ .
So, $P(A') = 1 - P(A)$ .

STEP 5

We know that $P(A) = \mathbf{0.4}$ , so $P(A') = 1 - \mathbf{0.4} = \mathbf{0.6}$ .
Awesome!

STEP 6

Now, we need to find $P(A' \cap B)$ , which means the probability of both $A'$ and $B$ happening.
Since $A$ and $B$ are independent, so are $A'$ and $B$ .
This means we can just multiply their probabilities!

STEP 7

So, $P(A' \cap B) = P(A') \cdot P(B)$ .
We just found $P(A') = \mathbf{0.6}$ and we know $P(B) = \mathbf{0.9}$ .

STEP 8

Let's multiply! $P(A' \cap B) = \mathbf{0.6} \cdot \mathbf{0.9} = \mathbf{0.54}$ .

STEP 9

We've got our final answer: $P(A' \cap B) = \mathbf{0.54}$ .
Which matches answer choice D!

STEP 10

The probability $P(A' \cap B)$ is 0.54, so the correct answer is D.

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QuestionSuppose $A$ and $B$ are independent events. If $P(A)=0.4$ and $P(B)=0.9$ , what is $P\left(A^{\prime} \cap B\right)$ ? A. 0.04 B. 0.36 C. 0.06 D. 0.54

Studdy Solution

STEP 1

What is this asking? If two events are independent, and we know how likely each one is to happen, how likely is it that the first one doesn't happen and the second one does? Watch out! Don't forget what "independent events" means, and don't mix up intersection and union!

STEP 2

1. Define Independence
2. Calculate the Probability of A'
3. Calculate the Probability of A' intersect B
4. Find the Answer

STEP 3

Alright, so, independent events mean that one event happening doesn't change the chance of the other event happening.
Think of flipping a coin and rolling a die – they don't affect each other!

STEP 4

$A'$ is the complement of $A$ , meaning it's everything except $A$ .
Since the total probability is always 1, the probability of $A'$ is $1$ minus the probability of $A$ .
So, $P(A') = 1 - P(A)$ .

STEP 5

We know that $P(A) = \mathbf{0.4}$ , so $P(A') = 1 - \mathbf{0.4} = \mathbf{0.6}$ .
Awesome!

STEP 6

Now, we need to find $P(A' \cap B)$ , which means the probability of both $A'$ and $B$ happening.
Since $A$ and $B$ are independent, so are $A'$ and $B$ .
This means we can just multiply their probabilities!

STEP 7

So, $P(A' \cap B) = P(A') \cdot P(B)$ .
We just found $P(A') = \mathbf{0.6}$ and we know $P(B) = \mathbf{0.9}$ .

STEP 8

Let's multiply! $P(A' \cap B) = \mathbf{0.6} \cdot \mathbf{0.9} = \mathbf{0.54}$ .

STEP 9

We've got our final answer: $P(A' \cap B) = \mathbf{0.54}$ .
Which matches answer choice D!

STEP 10

The probability $P(A' \cap B)$ is 0.54, so the correct answer is D.

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QuestionSuppose AAA and BBB are independent events. If P(A)=0.4P(A)=0.4P(A)=0.4 and P(B)=0.9P(B)=0.9P(B)=0.9, what is P(A′∩B)P\left(A^{\prime} \cap B\right)P(A′∩B) ? A. 0.04 B. 0.36 C. 0.06 D. 0.54

Studdy Solution

STEP 1

What is this asking? If two events are independent, and we know how likely each one is to happen, how likely is it that the first one *doesn't* happen and the second one *does*? Watch out! Don't forget what "independent events" means, and don't mix up intersection and union!

STEP 2

1. Define Independence2. Calculate the Probability of A'3. Calculate the Probability of A' intersect B4. Find the Answer

STEP 3

Alright, so, *independent events* mean that one event happening doesn't change the chance of the other event happening.Think of flipping a coin and rolling a die – they don't affect each other!

STEP 4

A′A'A′ is the *complement* of AAA, meaning it's everything *except* AAA.Since the total probability is always **1**, the probability of A′A'A′ is 111 minus the probability of AAA.So, P(A′)=1−P(A)P(A') = 1 - P(A)P(A′)=1−P(A).

STEP 5

We know that P(A)=0.4P(A) = \mathbf{0.4}P(A)=0.4, so P(A′)=1−0.4=0.6P(A') = 1 - \mathbf{0.4} = \mathbf{0.6}P(A′)=1−0.4=0.6.Awesome!

STEP 6

Now, we need to find P(A′∩B)P(A' \cap B)P(A′∩B), which means the probability of *both* A′A'A′ and BBB happening.Since AAA and BBB are independent, so are A′A'A′ and BBB.This means we can just multiply their probabilities!

STEP 7

So, P(A′∩B)=P(A′)⋅P(B)P(A' \cap B) = P(A') \cdot P(B)P(A′∩B)=P(A′)⋅P(B).We just found P(A′)=0.6P(A') = \mathbf{0.6}P(A′)=0.6 and we know P(B)=0.9P(B) = \mathbf{0.9}P(B)=0.9.

STEP 8

Let's multiply! P(A′∩B)=0.6⋅0.9=0.54P(A' \cap B) = \mathbf{0.6} \cdot \mathbf{0.9} = \mathbf{0.54}P(A′∩B)=0.6⋅0.9=0.54.

STEP 9

We've got our final answer: P(A′∩B)=0.54P(A' \cap B) = \mathbf{0.54}P(A′∩B)=0.54.Which matches answer choice **D**!

STEP 10

The probability P(A′∩B)P(A' \cap B)P(A′∩B) is **0.54**, so the correct answer is **D**.

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Studdy solves anything!

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QuestionSuppose $A$ and $B$ are independent events. If $P(A)=0.4$ and $P(B)=0.9$ , what is $P\left(A^{\prime} \cap B\right)$ ? A. 0.04 B. 0.36 C. 0.06 D. 0.54

What is this asking? If two events are independent, and we know how likely each one is to happen, how likely is it that the first one doesn't happen and the second one does? Watch out! Don't forget what "independent events" means, and don't mix up intersection and union!

1. Define Independence
2. Calculate the Probability of A'
3. Calculate the Probability of A' intersect B
4. Find the Answer

Alright, so, independent events mean that one event happening doesn't change the chance of the other event happening.
Think of flipping a coin and rolling a die – they don't affect each other!

$A'$ is the complement of $A$ , meaning it's everything except $A$ .
Since the total probability is always 1, the probability of $A'$ is $1$ minus the probability of $A$ .
So, $P(A') = 1 - P(A)$ .

We know that $P(A) = \mathbf{0.4}$ , so $P(A') = 1 - \mathbf{0.4} = \mathbf{0.6}$ .
Awesome!

Now, we need to find $P(A' \cap B)$ , which means the probability of both $A'$ and $B$ happening.
Since $A$ and $B$ are independent, so are $A'$ and $B$ .
This means we can just multiply their probabilities!

So, $P(A' \cap B) = P(A') \cdot P(B)$ .
We just found $P(A') = \mathbf{0.6}$ and we know $P(B) = \mathbf{0.9}$ .

Let's multiply! $P(A' \cap B) = \mathbf{0.6} \cdot \mathbf{0.9} = \mathbf{0.54}$ .

We've got our final answer: $P(A' \cap B) = \mathbf{0.54}$ .
Which matches answer choice D!

The probability $P(A' \cap B)$ is 0.54, so the correct answer is D.