Math / Data & Statistics

Question $P(B \mid D)=0.55$ $\begin{array}{ll} P(D)=0.4 \\ P(\bar{B} \mid D)=0.45 & P(D \cap \bar{B})=0.18 \\ P(\bar{D})=0.6] & P(\bar{D} \cap B)=0.35 \\ P(\bar{B} \mid \bar{D})=\square \end{array}$

Studdy Solution

STEP 1

What is this asking? We're given probabilities related to two events, B and D, and we need to find the probability of "not B" given "not D". Watch out! Don't mix up conditional probabilities with joint probabilities!
Also, remember the complement rule!

STEP 2

1. Find P(B and not D)
2. Find P(not D)
3. Calculate P(not B given not D)

STEP 3

We know $P(D) = 0.4$ , so the probability of "not D" is $P(\bar{D}) = 1 - P(D) = 1 - 0.4 = \mathbf{0.6}$ .
We're dividing to one here because the probabilities of an event and its complement must add up to one!

STEP 4

We are given $P(\bar{D} \cap B) = 0.35$ , which is exactly what we were looking for, the probability of B happening and D not happening.
Awesome!

STEP 5

We already found this in the first step! $P(\bar{D}) = \mathbf{0.6}$ .
Sometimes, the steps we need are already done!

STEP 6

The conditional probability formula is $P(A \mid B) = \frac{P(A \cap B)}{P(B)}$ .
In our case, we want $P(\bar{B} \mid \bar{D})$ , so we'll use $A = \bar{B}$ and $B = \bar{D}$ .

STEP 7

Plugging in the values we found, we get $P(\bar{B} \mid \bar{D}) = \frac{P(\bar{B} \cap \bar{D})}{P(\bar{D})}$ .

STEP 8

We know that $P(\bar{D} \cap B) = 0.35$ and $P(\bar{D}) = 0.6$ .
We need $P(\bar{B} \cap \bar{D})$ .
We aren't given that directly, but don't worry!

STEP 9

We know $P(B \mid D) = 0.55$ .
Since $P(B \mid D) + P(\bar{B} \mid D) = 1$ , we can find $P(\bar{B} \mid D) = 1 - 0.55 = 0.45$ .
This tells us that if D happens, B doesn't happen with a probability of 0.45.

STEP 10

We're given $P(D \cap \bar{B}) = 0.18$ .
This is the probability that D happens and B doesn't.

STEP 11

We're given $P(\bar{D} \cap B) = 0.35$ .
This is the probability that D doesn't happen and B does.

STEP 12

We know that $P(B) = P(B \cap D) + P(B \cap \bar{D})$ .
Since $P(B \cap \bar{D}) = 0.35$ , we just need $P(B \cap D)$ .
We know $P(B \mid D) = \frac{P(B \cap D)}{P(D)}$ .
We have $P(B \mid D) = 0.55$ and $P(D) = 0.4$ , so $P(B \cap D) = 0.55 \cdot 0.4 = 0.22$ .

STEP 13

Now we can find $P(B) = P(B \cap D) + P(B \cap \bar{D}) = 0.22 + 0.35 = 0.57$ .
Then, $P(\bar{B}) = 1 - P(B) = 1 - 0.57 = 0.43$ .

STEP 14

We also know $P(\bar{B}) = P(\bar{B} \cap D) + P(\bar{B} \cap \bar{D})$ .
We have $P(\bar{B}) = 0.43$ and $P(\bar{B} \cap D) = 0.18$ , so $P(\bar{B} \cap \bar{D}) = 0.43 - 0.18 = 0.25$ .

STEP 15

Finally! $P(\bar{B} \mid \bar{D}) = \frac{P(\bar{B} \cap \bar{D})}{P(\bar{D})} = \frac{0.25}{0.6} = \frac{5}{12} \approx 0.4167$ .

STEP 16

$P(\bar{B} \mid \bar{D}) = \frac{5}{12}$

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Question $P(B \mid D)=0.55$ $\begin{array}{ll} P(D)=0.4 \\ P(\bar{B} \mid D)=0.45 & P(D \cap \bar{B})=0.18 \\ P(\bar{D})=0.6] & P(\bar{D} \cap B)=0.35 \\ P(\bar{B} \mid \bar{D})=\square \end{array}$

Studdy Solution

STEP 1

What is this asking? We're given probabilities related to two events, B and D, and we need to find the probability of "not B" given "not D". Watch out! Don't mix up conditional probabilities with joint probabilities!
Also, remember the complement rule!

STEP 2

1. Find P(B and not D)
2. Find P(not D)
3. Calculate P(not B given not D)

STEP 3

We know $P(D) = 0.4$ , so the probability of "not D" is $P(\bar{D}) = 1 - P(D) = 1 - 0.4 = \mathbf{0.6}$ .
We're dividing to one here because the probabilities of an event and its complement must add up to one!

STEP 4

We are given $P(\bar{D} \cap B) = 0.35$ , which is exactly what we were looking for, the probability of B happening and D not happening.
Awesome!

STEP 5

We already found this in the first step! $P(\bar{D}) = \mathbf{0.6}$ .
Sometimes, the steps we need are already done!

STEP 6

The conditional probability formula is $P(A \mid B) = \frac{P(A \cap B)}{P(B)}$ .
In our case, we want $P(\bar{B} \mid \bar{D})$ , so we'll use $A = \bar{B}$ and $B = \bar{D}$ .

STEP 7

Plugging in the values we found, we get $P(\bar{B} \mid \bar{D}) = \frac{P(\bar{B} \cap \bar{D})}{P(\bar{D})}$ .

STEP 8

We know that $P(\bar{D} \cap B) = 0.35$ and $P(\bar{D}) = 0.6$ .
We need $P(\bar{B} \cap \bar{D})$ .
We aren't given that directly, but don't worry!

STEP 9

We know $P(B \mid D) = 0.55$ .
Since $P(B \mid D) + P(\bar{B} \mid D) = 1$ , we can find $P(\bar{B} \mid D) = 1 - 0.55 = 0.45$ .
This tells us that if D happens, B doesn't happen with a probability of 0.45.

STEP 10

We're given $P(D \cap \bar{B}) = 0.18$ .
This is the probability that D happens and B doesn't.

STEP 11

We're given $P(\bar{D} \cap B) = 0.35$ .
This is the probability that D doesn't happen and B does.

STEP 12

We know that $P(B) = P(B \cap D) + P(B \cap \bar{D})$ .
Since $P(B \cap \bar{D}) = 0.35$ , we just need $P(B \cap D)$ .
We know $P(B \mid D) = \frac{P(B \cap D)}{P(D)}$ .
We have $P(B \mid D) = 0.55$ and $P(D) = 0.4$ , so $P(B \cap D) = 0.55 \cdot 0.4 = 0.22$ .

STEP 13

Now we can find $P(B) = P(B \cap D) + P(B \cap \bar{D}) = 0.22 + 0.35 = 0.57$ .
Then, $P(\bar{B}) = 1 - P(B) = 1 - 0.57 = 0.43$ .

STEP 14

We also know $P(\bar{B}) = P(\bar{B} \cap D) + P(\bar{B} \cap \bar{D})$ .
We have $P(\bar{B}) = 0.43$ and $P(\bar{B} \cap D) = 0.18$ , so $P(\bar{B} \cap \bar{D}) = 0.43 - 0.18 = 0.25$ .

STEP 15

Finally! $P(\bar{B} \mid \bar{D}) = \frac{P(\bar{B} \cap \bar{D})}{P(\bar{D})} = \frac{0.25}{0.6} = \frac{5}{12} \approx 0.4167$ .

STEP 16

$P(\bar{B} \mid \bar{D}) = \frac{5}{12}$

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Studdy Solution

STEP 1

What is this asking? We're given probabilities related to two events, B and D, and we need to find the probability of "not B" given "not D". Watch out! Don't mix up conditional probabilities with joint probabilities!Also, remember the complement rule!

STEP 2

1. Find P(B and not D)2. Find P(not D)3. Calculate P(not B given not D)

STEP 3

We know P(D)=0.4P(D) = 0.4P(D)=0.4, so the probability of "not D" is P(Dˉ)=1−P(D)=1−0.4=0.6P(\bar{D}) = 1 - P(D) = 1 - 0.4 = \mathbf{0.6}P(Dˉ)=1−P(D)=1−0.4=0.6.We're dividing to one here because the probabilities of an event and its complement must add up to one!

STEP 4

We are given P(Dˉ∩B)=0.35P(\bar{D} \cap B) = 0.35P(Dˉ∩B)=0.35, which is exactly what we were looking for, the probability of B happening *and* D not happening.Awesome!

STEP 5

We already found this in the first step! P(Dˉ)=0.6P(\bar{D}) = \mathbf{0.6}P(Dˉ)=0.6.Sometimes, the steps we need are already done!

STEP 6

The **conditional probability** formula is 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)​.In our case, we want P(Bˉ∣Dˉ)P(\bar{B} \mid \bar{D})P(Bˉ∣Dˉ), so we'll use A=BˉA = \bar{B}A=Bˉ and B=DˉB = \bar{D}B=Dˉ.

STEP 7

Plugging in the values we found, we get P(Bˉ∣Dˉ)=P(Bˉ∩Dˉ)P(Dˉ)P(\bar{B} \mid \bar{D}) = \frac{P(\bar{B} \cap \bar{D})}{P(\bar{D})}P(Bˉ∣Dˉ)=P(Dˉ)P(Bˉ∩Dˉ)​.

STEP 8

We know that P(Dˉ∩B)=0.35P(\bar{D} \cap B) = 0.35P(Dˉ∩B)=0.35 and P(Dˉ)=0.6P(\bar{D}) = 0.6P(Dˉ)=0.6.We need P(Bˉ∩Dˉ)P(\bar{B} \cap \bar{D})P(Bˉ∩Dˉ).We aren't given that directly, but don't worry!

STEP 9

STEP 10

We're given P(D∩Bˉ)=0.18P(D \cap \bar{B}) = 0.18P(D∩Bˉ)=0.18.This is the probability that D happens *and* B doesn't.

STEP 11

We're given P(Dˉ∩B)=0.35P(\bar{D} \cap B) = 0.35P(Dˉ∩B)=0.35.This is the probability that D *doesn't* happen and B *does*.

STEP 12

STEP 13

Now we can find P(B)=P(B∩D)+P(B∩Dˉ)=0.22+0.35=0.57P(B) = P(B \cap D) + P(B \cap \bar{D}) = 0.22 + 0.35 = 0.57P(B)=P(B∩D)+P(B∩Dˉ)=0.22+0.35=0.57.Then, P(Bˉ)=1−P(B)=1−0.57=0.43P(\bar{B}) = 1 - P(B) = 1 - 0.57 = 0.43P(Bˉ)=1−P(B)=1−0.57=0.43.

STEP 14

STEP 15

Finally! P(Bˉ∣Dˉ)=P(Bˉ∩Dˉ)P(Dˉ)=0.250.6=512≈0.4167P(\bar{B} \mid \bar{D}) = \frac{P(\bar{B} \cap \bar{D})}{P(\bar{D})} = \frac{0.25}{0.6} = \frac{5}{12} \approx 0.4167P(Bˉ∣Dˉ)=P(Dˉ)P(Bˉ∩Dˉ)​=0.60.25​=125​≈0.4167.

STEP 16

P(Bˉ∣Dˉ)=512P(\bar{B} \mid \bar{D}) = \frac{5}{12}P(Bˉ∣Dˉ)=125​

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Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

What is this asking? We're given probabilities related to two events, B and D, and we need to find the probability of "not B" given "not D". Watch out! Don't mix up conditional probabilities with joint probabilities!
Also, remember the complement rule!

1. Find P(B and not D)
2. Find P(not D)
3. Calculate P(not B given not D)

We know $P(D) = 0.4$ , so the probability of "not D" is $P(\bar{D}) = 1 - P(D) = 1 - 0.4 = \mathbf{0.6}$ .
We're dividing to one here because the probabilities of an event and its complement must add up to one!

We are given $P(\bar{D} \cap B) = 0.35$ , which is exactly what we were looking for, the probability of B happening and D not happening.
Awesome!

We already found this in the first step! $P(\bar{D}) = \mathbf{0.6}$ .
Sometimes, the steps we need are already done!

The conditional probability formula is $P(A \mid B) = \frac{P(A \cap B)}{P(B)}$ .
In our case, we want $P(\bar{B} \mid \bar{D})$ , so we'll use $A = \bar{B}$ and $B = \bar{D}$ .

Plugging in the values we found, we get $P(\bar{B} \mid \bar{D}) = \frac{P(\bar{B} \cap \bar{D})}{P(\bar{D})}$ .

We know that $P(\bar{D} \cap B) = 0.35$ and $P(\bar{D}) = 0.6$ .
We need $P(\bar{B} \cap \bar{D})$ .
We aren't given that directly, but don't worry!

We're given $P(D \cap \bar{B}) = 0.18$ .
This is the probability that D happens and B doesn't.

We're given $P(\bar{D} \cap B) = 0.35$ .
This is the probability that D doesn't happen and B does.

Now we can find $P(B) = P(B \cap D) + P(B \cap \bar{D}) = 0.22 + 0.35 = 0.57$ .
Then, $P(\bar{B}) = 1 - P(B) = 1 - 0.57 = 0.43$ .

Finally! $P(\bar{B} \mid \bar{D}) = \frac{P(\bar{B} \cap \bar{D})}{P(\bar{D})} = \frac{0.25}{0.6} = \frac{5}{12} \approx 0.4167$ .

$P(\bar{B} \mid \bar{D}) = \frac{5}{12}$