Answer:
300
Step-by-step explanation:
Answer:
What is the probability that a randomly selected family owns a cat? 34%
What is the conditional probability that a randomly selected family doesn't own a dog given that it owns a cat? 82.4%
Step-by-step explanation: We can use a Venn (attached) diagram to describe this situation:
Imagine a community of 100 families (we can assum a number, because in the end, it does not matter)
So, 30% of the families own a dog = .30*100 = 30
20% of the families that own a dog also own a cat = 0.2*30 = 6
34% of all the families own a cat = 0.34*100 = 34
Dogs and cats: 6
Only dogs: 30 - 6 = 24
Only cats: 34 - 6 = 28
Not cat and dogs: 24+6+28 = 58; 100 - 58 = 42
What is the probability that a randomly selected family owns a cat?
34/100 = 34%
What is the conditional probability that a randomly selected family doesn't own a dog given that it owns a cat?
A = doesn't own a dog
B = owns a cat
P(A|B) = P(A∩B)/P(B) = 28/34 = 82.4%
Answer:
Step-by-step explanation:
Question (1).
OQ and RT are the parallel lines and UN is a transversal intersecting these lines at two different points P and S.
A). ∠OPS ≅ ∠RSU [corresponding angles]
B). m∠OPS + m∠RSP = 180° [Consecutive interior angles]
C). m∠OPS + m∠OPN = 180° [Linear pair of angles]
D). Since, ∠OPS ≅ ∠TSP [Alternate interior angles]
And m∠TSP + m∠TSU = 180° [Linear pair of angles]
Therefore, Option (A) is the correct option.
Question (2).
A). m∠RSP + m∠RSU = 180° [Linear pair of angles]
B). m∠RSP + m∠PST = 180° [Linear pair of angles]
C). ∠RSP ≅ ∠TSU [Vertically opposite angles]
D). m∠RSP + m∠OPS = 180° [Consecutive interior angles]
Therefore, Option (C) will be the answer.
Answer:
Step-by-step explanation:
Given that events A1, A2 and A3 form a partiton of the sample space S with probabilities P(A1) = 0.3, P(A2) = 0.5, P(A3) = 0.2.
i.e. A1, A2, and A3 are mutually exclusive and exhaustive
E is an event such that
P(E|A1) = 0.1, P(E|A2) = 0.6, P(E|A3) = 0.8,



