Answer:
a) 0.06
b) 0.778
Step-by-step explanation:
Let's suppose a community of 100 families just to facilitate the calculation.
30% of the families own a dog
Dog = 30% of 100 = 30
20% of the families that own a dog also own a cat = 20% of 30 = 6
27% of all the families own a cat = 27% of 100 = 27
So, 6 families own a dog and a cat.
As 30 families own a dog, [30 - 6 =] 24 families own only dogs
As 27 families own a cat, [27 - 6 = ] 21 families own only cats
See picture attached.
a) What is the probability that a randomly selected family owns both a dog and a cat?
P(dog and cat) = dog ∩ cat/total = 6/100 = 0.06
b) What is the conditional probability that a randomly selected family doesn't own a dog given that it owns a cat?
So, only cat/total cat
P (not dog|cat) = 21/27 = 0.778
1/4 of 12 is 3. Because there are 4 threes in twelve. 12/4 is twelve divided by four, which is three.
You multiply 12.5 points a game by 24 games and get 300 points for a season
The events A and B are independent events, and the values of P(A) and P(B) are 7/12 and 1/2, respectively
<h3>The value of P(A)</h3>
The event A is given as:
A : Sum greater than 6
In the sample space of a roll of two dice, there are 21 outcomes that are greater than 6, out of a total of 36 outcomes
This means that:
P(A) = 21/36
Simplify
P(A) = 7/12
<h3>The value of P(B)</h3>
The event B is given as:
B : Sum is divisible by 2
In the sample space of a roll of two dice, there are 18 outcomes that are divisible by 2, out of a total of 36 outcomes
This means that:
P(B) = 18/36
Simplify
P(B) = 1/2
Hence, the probability values of P(A) and P(B) are 7/12 and 1/2, respectively
Read more about probability at:
brainly.com/question/251701
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Answer:
m = -10
Step-by-step explanation:
>:)
