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In a study of canine attacks, the probability that the victim was under 18 years of age is 80%. the probability that the attack occurred on the dog owner's property was 64%. the probability that the victim was under 18 years of age or the attack occurred on the dog owner's propery was 95%. what is the probability that the victim was under 18 years of age and the attack occurred on the dog owner's property?

Sagot :

The probability that the victim's age of a canine attack was under 18 years and the attack took place at the property of the dog owner is 49/100

Let A be the event of the canine attack victim being under 18 years of age.

Let B be the event of the canine attack occurring at the dog owner's property.

It is given that-

P(A) = 80%

= 80/100

= 4/5

P(B) = 64%

= 64/100

= 16/25

It is also given that

P(A or B) = 95%

or, P(A ∪ B) = 95%

= 95/100

= 19/20

We need to find the probability that the victim was aged under 18 years and that the attack occurred on the dog owner's property.

= P(A and B) = P(A ∩ B)

We know that

P(X ∪ Y) = P(X) + P(Y) - P(X ∩ Y)

or, P(X ∩ Y) = P(X) + P(Y) - P(X ∪ Y)

Therefore,

P(A ∩ B) = P(A) + P(B) - P(A ∪ B)

= 4/5 + 16/25 - 19/20

taking LCM of denominators we get

= (80 + 64 - 95)/100

= 49/100

To learn more about probability visit

https://brainly.com/question/14603083

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