Math / Data & Statistics

QuestionThe table gives a set of outcomes and their probabilities. Let $A$ be the event "the outcome is divisible by $2 "$ . Let $B$ be the event "the outcome is prime". Find $P(A \mid B)$ . \begin{tabular}{|c|c|} \hline Outcome & Probability \\ \hline 1 & 0.3 \\ \hline 2 & 0.4 \\ \hline 3 & 0.1 \\ \hline 4 & 0.2 \\ \hline \end{tabular} $\square$

Studdy Solution

STEP 1

What is this asking? If we KNOW that a randomly picked outcome is a prime number, what's the chance it's also even? Watch out! Don't mix up $P(A \mid B)$ and $P(B \mid A)$ !
We're given that B has already happened, not A.

STEP 2

1. Find the prime outcomes
2. Calculate $P(B)$
3. Find the outcomes in both A and B
4. Calculate $P(A \cap B)$
5. Calculate $P(A \mid B)$

STEP 3

Let's list out the possible outcomes and their probabilities!
We have $1$ with $0.3$ , $2$ with $0.4$ , $3$ with $0.1$ , and $4$ with $0.2$ .

STEP 4

Remember, prime numbers are whole numbers greater than $1$ that are only divisible by $1$ and themselves.
Looking at our outcomes, the prime numbers are $2$ and $3$ .

STEP 5

$P(B)$ , the probability of event $B$ , is the probability that the outcome is prime.
Since $2$ and $3$ are our prime outcomes, we add their probabilities: $P(B) = P(2) + P(3)$ .

STEP 6

Substituting the values from our table, we get $P(B) = 0.4 + 0.1 = \mathbf{0.5}$ .
So, there's a 50% chance of getting a prime outcome!

STEP 7

Event $A$ is "the outcome is divisible by $2$ ".
We need to find outcomes that are both prime AND divisible by $2$ .

STEP 8

Looking at our prime outcomes ( $2$ and $3$ ), only $2$ is divisible by $2$ .
So, the outcome that satisfies both $A$ and $B$ is $2$ .

STEP 9

$P(A \cap B)$ is the probability that both $A$ and $B$ occur.
Since the only outcome in both $A$ and $B$ is $2$ , $P(A \cap B)$ is simply the probability of getting a $2$ .

STEP 10

From the table, the probability of getting a $2$ is $0.4$ .
So, $P(A \cap B) = \mathbf{0.4}$ .

STEP 11

Finally, we can calculate $P(A \mid B)$ using the formula: $P(A \mid B) = \frac{P(A \cap B)}{P(B)}$ This formula tells us the probability of $A$ happening given that $B$ has already happened.

STEP 12

We already found $P(A \cap B) = 0.4$ and $P(B) = 0.5$ .
Plugging these values into the formula, we get: $P(A \mid B) = \frac{0.4}{0.5} = \frac{4}{5} = \mathbf{0.8}$

STEP 13

$P(A \mid B) = 0.8$ .
There's an 80% chance that the outcome is divisible by $2$ given that it's prime!

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Studdy Solution

STEP 1

What is this asking? If we KNOW that a randomly picked outcome is a prime number, what's the chance it's also even? Watch out! Don't mix up $P(A \mid B)$ and $P(B \mid A)$ !
We're given that B has already happened, not A.

STEP 2

1. Find the prime outcomes
2. Calculate $P(B)$
3. Find the outcomes in both A and B
4. Calculate $P(A \cap B)$
5. Calculate $P(A \mid B)$

STEP 3

Let's list out the possible outcomes and their probabilities!
We have $1$ with $0.3$ , $2$ with $0.4$ , $3$ with $0.1$ , and $4$ with $0.2$ .

STEP 4

Remember, prime numbers are whole numbers greater than $1$ that are only divisible by $1$ and themselves.
Looking at our outcomes, the prime numbers are $2$ and $3$ .

STEP 5

$P(B)$ , the probability of event $B$ , is the probability that the outcome is prime.
Since $2$ and $3$ are our prime outcomes, we add their probabilities: $P(B) = P(2) + P(3)$ .

STEP 6

Substituting the values from our table, we get $P(B) = 0.4 + 0.1 = \mathbf{0.5}$ .
So, there's a 50% chance of getting a prime outcome!

STEP 7

Event $A$ is "the outcome is divisible by $2$ ".
We need to find outcomes that are both prime AND divisible by $2$ .

STEP 8

Looking at our prime outcomes ( $2$ and $3$ ), only $2$ is divisible by $2$ .
So, the outcome that satisfies both $A$ and $B$ is $2$ .

STEP 9

$P(A \cap B)$ is the probability that both $A$ and $B$ occur.
Since the only outcome in both $A$ and $B$ is $2$ , $P(A \cap B)$ is simply the probability of getting a $2$ .

STEP 10

From the table, the probability of getting a $2$ is $0.4$ .
So, $P(A \cap B) = \mathbf{0.4}$ .

STEP 11

Finally, we can calculate $P(A \mid B)$ using the formula: $P(A \mid B) = \frac{P(A \cap B)}{P(B)}$ This formula tells us the probability of $A$ happening given that $B$ has already happened.

STEP 12

We already found $P(A \cap B) = 0.4$ and $P(B) = 0.5$ .
Plugging these values into the formula, we get: $P(A \mid B) = \frac{0.4}{0.5} = \frac{4}{5} = \mathbf{0.8}$

STEP 13

$P(A \mid B) = 0.8$ .
There's an 80% chance that the outcome is divisible by $2$ given that it's prime!

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Studdy Solution

STEP 1

What is this asking? If we KNOW that a randomly picked outcome is a **prime number**, what's the chance it's also **even**? Watch out! Don't mix up P(A∣B)P(A \mid B)P(A∣B) and P(B∣A)P(B \mid A)P(B∣A)!We're given that *B* has already happened, not *A*.

STEP 2

1. Find the prime outcomes2. Calculate P(B)P(B)P(B)3. Find the outcomes in both A and B4. Calculate P(A∩B)P(A \cap B)P(A∩B)5. Calculate P(A∣B)P(A \mid B)P(A∣B)

STEP 3

Let's **list out** the possible outcomes and their probabilities!We have 111 with 0.30.30.3, 222 with 0.40.40.4, 333 with 0.10.10.1, and 444 with 0.20.20.2.

STEP 4

Remember, **prime numbers** are whole numbers greater than 111 that are only divisible by 111 and themselves.Looking at our outcomes, the prime numbers are 222 and 333.

STEP 5

P(B)P(B)P(B), the probability of event BBB, is the probability that the outcome is **prime**.Since 222 and 333 are our prime outcomes, we **add their probabilities**: P(B)=P(2)+P(3)P(B) = P(2) + P(3)P(B)=P(2)+P(3).

STEP 6

Substituting the values from our table, we get P(B)=0.4+0.1=0.5P(B) = 0.4 + 0.1 = \mathbf{0.5}P(B)=0.4+0.1=0.5.So, there's a **50% chance** of getting a prime outcome!

STEP 7

Event AAA is "the outcome is divisible by 222".We need to find outcomes that are **both prime AND divisible by** 222.

STEP 8

Looking at our prime outcomes (222 and 333), only 222 is divisible by 222.So, the outcome that satisfies both AAA and BBB is 222.

STEP 9

P(A∩B)P(A \cap B)P(A∩B) is the probability that **both** AAA and BBB occur.Since the only outcome in both AAA and BBB is 222, P(A∩B)P(A \cap B)P(A∩B) is simply the probability of getting a 222.

STEP 10

From the table, the probability of getting a 222 is 0.40.40.4.So, P(A∩B)=0.4P(A \cap B) = \mathbf{0.4}P(A∩B)=0.4.

STEP 11

Finally, we can calculate P(A∣B)P(A \mid B)P(A∣B) using the formula: P(A∣B)=P(A∩B)P(B) P(A \mid B) = \frac{P(A \cap B)}{P(B)} P(A∣B)=P(B)P(A∩B)​ This formula tells us the probability of AAA happening *given* that BBB has already happened.

STEP 12

We already found P(A∩B)=0.4P(A \cap B) = 0.4P(A∩B)=0.4 and P(B)=0.5P(B) = 0.5P(B)=0.5.Plugging these values into the formula, we get: P(A∣B)=0.40.5=45=0.8 P(A \mid B) = \frac{0.4}{0.5} = \frac{4}{5} = \mathbf{0.8} P(A∣B)=0.50.4​=54​=0.8

STEP 13

P(A∣B)=0.8P(A \mid B) = 0.8P(A∣B)=0.8.There's an **80% chance** that the outcome is divisible by 222 given that it's prime!

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What is this asking? If we KNOW that a randomly picked outcome is a prime number, what's the chance it's also even? Watch out! Don't mix up $P(A \mid B)$ and $P(B \mid A)$ !
We're given that B has already happened, not A.

1. Find the prime outcomes
2. Calculate $P(B)$
3. Find the outcomes in both A and B
4. Calculate $P(A \cap B)$
5. Calculate $P(A \mid B)$

Let's list out the possible outcomes and their probabilities!
We have $1$ with $0.3$ , $2$ with $0.4$ , $3$ with $0.1$ , and $4$ with $0.2$ .

Remember, prime numbers are whole numbers greater than $1$ that are only divisible by $1$ and themselves.
Looking at our outcomes, the prime numbers are $2$ and $3$ .

$P(B)$ , the probability of event $B$ , is the probability that the outcome is prime.
Since $2$ and $3$ are our prime outcomes, we add their probabilities: $P(B) = P(2) + P(3)$ .

Substituting the values from our table, we get $P(B) = 0.4 + 0.1 = \mathbf{0.5}$ .
So, there's a 50% chance of getting a prime outcome!

Event $A$ is "the outcome is divisible by $2$ ".
We need to find outcomes that are both prime AND divisible by $2$ .

Looking at our prime outcomes ( $2$ and $3$ ), only $2$ is divisible by $2$ .
So, the outcome that satisfies both $A$ and $B$ is $2$ .

$P(A \cap B)$ is the probability that both $A$ and $B$ occur.
Since the only outcome in both $A$ and $B$ is $2$ , $P(A \cap B)$ is simply the probability of getting a $2$ .

From the table, the probability of getting a $2$ is $0.4$ .
So, $P(A \cap B) = \mathbf{0.4}$ .

Finally, we can calculate $P(A \mid B)$ using the formula: $P(A \mid B) = \frac{P(A \cap B)}{P(B)}$ This formula tells us the probability of $A$ happening given that $B$ has already happened.

We already found $P(A \cap B) = 0.4$ and $P(B) = 0.5$ .
Plugging these values into the formula, we get: $P(A \mid B) = \frac{0.4}{0.5} = \frac{4}{5} = \mathbf{0.8}$

$P(A \mid B) = 0.8$ .
There's an 80% chance that the outcome is divisible by $2$ given that it's prime!