In a group of 100 people 40 own a cat, 25 own a dog and 15 own a cat and a dog.
Find the probability that a person selected at random from the group:
a) owns a dog or a cat.
b) owns a dog or a cat, but not both
c) owns a dog, given that he owns a cat
d) does not own a cat, given that he owns a dog.
N(C or D) = N(C) + N(D) - N(C and D)
= 40 + 25 - 15 = 50
prob(C or D) = 50/100 = 1/2
N(Cat or Dog, not both) = 40 + 25 - 15 -15 = 35
Prob(that event) = 35100 = 7/20
c)
Prob( A | B) ----- conditional prob
= Prob( A and B)/Prob(b)
prob(dog | cat) = prob(dog and cat)/prob(cat)
= (15/100)/(40/100) = 3/8
d)
prob(cat | dog) = prob(cat and dog)/prob(dog)
= (15/100) / (25/100) = 3/5
a) The probability that a person selected at random from the group owns a dog or a cat can be calculated as follows:
P(owns a dog or a cat) = P(owns a dog) + P(owns a cat) - P(owns a dog and a cat)
P(owns a dog) = 25/100 = 0.25
P(owns a cat) = 40/100 = 0.4
P(owns a dog and a cat) = 15/100 = 0.15
P(owns a dog or a cat) = 0.25 + 0.4 - 0.15 = 0.5
Therefore, the probability that a person selected at random from the group owns a dog or a cat is 0.5.
b) The probability that a person selected at random from the group owns a dog or a cat, but not both can be calculated as follows:
P(owns a dog or a cat, but not both) = P(owns a dog) + P(owns a cat) - 2 * P(owns a dog and a cat)
Using the previous values:
P(owns a dog or a cat, but not both) = 0.25 + 0.4 - 2 * 0.15 = 0.6 - 0.3 = 0.3
Therefore, the probability that a person selected at random from the group owns a dog or a cat, but not both is 0.3.
c) The probability that a person selected at random from the group owns a dog, given that he owns a cat can be calculated as follows:
P(owns a dog | owns a cat) = P(owns a dog and a cat) / P(owns a cat)
Using the previous values:
P(owns a dog | owns a cat) = 0.15 / 0.4 = 0.375
Therefore, the probability that a person selected at random from the group owns a dog, given that he owns a cat is 0.375.
d) The probability that a person selected at random from the group does not own a cat, given that he owns a dog can be calculated as follows:
P(does not own a cat | owns a dog) = P(owns a dog) - P(owns a dog and a cat) / P(owns a dog)
Using the previous values:
P(does not own a cat | owns a dog) = 0.25 - 0.15 / 0.25 = 0.1 / 0.25 = 0.4
Therefore, the probability that a person selected at random from the group does not own a cat, given that he owns a dog is 0.4.
Given information:
- Total number of people in the group (N) = 100
- Number of people who own a cat (C) = 40
- Number of people who own a dog (D) = 25
- Number of people who own both a cat and a dog (C ∩ D) = 15
a) To find the probability that a person selected at random from the group owns a dog or a cat, we can use the principle of inclusion-exclusion.
P(D or C) = P(D) + P(C) - P(D ∩ C)
In this case, P(D) is the probability of owning a dog, P(C) is the probability of owning a cat, and P(D ∩ C) is the probability of owning both a cat and a dog.
P(D or C) = (Number of people who own a dog) / N + (Number of people who own a cat) / N - (Number of people who own both a cat and a dog) / N
P(D or C) = D / N + C / N - (C ∩ D) / N
P(D or C) = 25/100 + 40/100 - 15/100
P(D or C) = 50/100 = 0.5 (or 50%)
Therefore, the probability that a person selected at random from the group owns a dog or a cat is 0.5 or 50%.
b) To find the probability that a person selected at random from the group owns a dog or a cat, but not both, we need to subtract the probability of owning both a cat and a dog from the probability of owning a dog or a cat.
P((D or C) but not (D ∩ C)) = P(D or C) - P(D ∩ C)
P((D or C) but not (D ∩ C)) = 0.5 - 0.15
P((D or C) but not (D ∩ C)) = 0.35 (or 35%)
Therefore, the probability that a person selected at random from the group owns a dog or a cat, but not both, is 0.35 or 35%.
c) To find the probability that a person selected at random from the group owns a dog given that they own a cat, we can use conditional probability.
P(D | C) = P(D ∩ C) / P(C)
In this case, P(D ∩ C) is the probability of owning both a cat and a dog, and P(C) is the probability of owning a cat.
P(D | C) = (Number of people who own both a cat and a dog) / (Number of people who own a cat)
P(D | C) = (C ∩ D) / C
P(D | C) = 15/40
P(D | C) = 0.375 (or 37.5%)
Therefore, the probability that a person selected at random from the group owns a dog, given that they own a cat, is 0.375 or 37.5%.
d) To find the probability that a person selected at random from the group does not own a cat, given that they own a dog, we can use conditional probability.
P(not C | D) = P(not C ∩ D) / P(D)
In this case, P(not C ∩ D) is the probability of owning a dog but not a cat, and P(D) is the probability of owning a dog.
P(not C | D) = (Number of people who own a dog but not a cat) / (Number of people who own a dog)
P(not C | D) = (D - C ∩ D) / D
P(not C | D) = (25 - 15) / 25
P(not C | D) = 0.4 (or 40%)
Therefore, the probability that a person selected at random from the group does not own a cat, given that they own a dog, is 0.4 or 40%.
To answer these questions, we can use the concept of set theory and the principle of inclusion-exclusion.
Let's denote:
A = the set of people who own a cat
B = the set of people who own a dog
a) To find the probability that a person selected at random owns a dog or a cat, we need to find the union of sets A and B, denoted by A ∪ B. In other words, we need to find the number of people who own either a cat or a dog.
To calculate this, we can use the formula:
|A ∪ B| = |A| + |B| - |A ∩ B|,
where |A ∪ B| represents the cardinality (number of elements) of A ∪ B, |A| represents the cardinality of A, |B| represents the cardinality of B, and |A ∩ B| represents the cardinality of the intersection of A and B.
Given that 40 people own a cat, 25 people own a dog, and 15 people own both a cat and a dog, we have:
|A| = 40,
|B| = 25,
|A ∩ B| = 15.
Substituting these values into the formula, we have:
|A ∪ B| = 40 + 25 - 15 = 50.
Therefore, the probability that a person selected at random from the group owns a dog or a cat is |A ∪ B| / 100 = 50 / 100 = 0.5 (or 50%).
b) To find the probability that a person selected at random owns a dog or a cat, but not both, we need to exclude the people who own both a cat and a dog from the previous calculation. In other words, we need to find the number of people who own either a cat or a dog, but not both.
|A ∪ B| - |A ∩ B| = 50 - 15 = 35.
Therefore, the probability that a person selected at random from the group owns a dog or a cat, but not both, is 35 / 100 = 0.35 (or 35%).
c) To find the probability that a person owns a dog, given that he owns a cat, we need to calculate the conditional probability P(B|A), which is the probability of owning a dog given that the person owns a cat.
The formula for conditional probability is:
P(B|A) = P(A ∩ B) / P(A).
We are given that 15 people own both a cat and a dog (|A ∩ B| = 15) and 40 people own a cat (|A| = 40).
Substituting these values into the formula, we have:
P(B|A) = 15 / 40 = 0.375 (or 37.5%).
Therefore, the probability that a person selected at random from the group owns a dog, given that he owns a cat, is 0.375 (or 37.5%).
d) To find the probability that a person does not own a cat, given that he owns a dog, we need to calculate the conditional probability P(~A|B), which is the probability of not owning a cat given that the person owns a dog.
The formula for conditional probability is:
P(~A|B) = P(B ∩ ~A) / P(B).
Since we know that P(B ∩ ~A) = P(B) - P(A ∩ B), we can calculate it by subtracting the probability of owning both a cat and a dog (15) from the probability of owning a dog (25).
P(B ∩ ~A) = 25 - 15 = 10.
Given that 25 people own a dog (|B| = 25), we can calculate the conditional probability:
P(~A|B) = P(B ∩ ~A) / P(B) = 10 / 25 = 0.4 (or 40%).
Therefore, the probability that a person selected at random from the group does not own a cat, given that he owns a dog, is 0.4 (or 40%).