In a class, the probability that a randomly chosen student likes dogs is 0.72, that a student likes cats is 0.54, and that a student likes neither is 0.11. Find the probability that a student likes both dogs and cats.
.72+.54-x = 1-.11
Pentagon MNOPQ is inscribed in a circle, NO=OP=PQ and MN=MQ. If angle NMQ=96°, find angle NPM.
If the center is C, then we know that angles
NCO = OCP = PCQ = 64°
because each subtends an equal side, and their sum is 2*96°.
Se where that takes you.
To find the probability that a student likes both dogs and cats, we need to consider the probabilities of liking dogs, liking cats, and liking neither.
Let's assume:
- P(D) represents the probability of liking dogs.
- P(C) represents the probability of liking cats.
- P(N) represents the probability of liking neither dogs nor cats.
- P(B) represents the probability of liking both dogs and cats.
We are given:
P(D) = 0.72 (probability of liking dogs)
P(C) = 0.54 (probability of liking cats)
P(N) = 0.11 (probability of liking neither dogs nor cats)
We know that the total probability of any event happening is 1. Therefore, we can say:
P(D) + P(C) + P(N) = 1
We need to find P(B) (probability of liking both dogs and cats).
We can use the Principle of Inclusion-Exclusion to find P(B):
P(B) = P(D) + P(C) - P(D ∪ C)
Here, P(D ∪ C) represents the probability of liking either dogs or cats or both. In other words, it represents the probability of liking at least one of them.
To find P(D ∪ C), we subtract P(N) from 1, as P(N) represents the probability of liking neither dogs nor cats:
P(D ∪ C) = 1 - P(N)
Substituting the given values, we can calculate P(D ∪ C):
P(D ∪ C) = 1 - P(N)
P(D ∪ C) = 1 - 0.11
P(D ∪ C) = 0.89
Now, substituting the calculated values back into the equation for P(B), we can find the probability of liking both dogs and cats:
P(B) = P(D) + P(C) - P(D ∪ C)
P(B) = 0.72 + 0.54 - 0.89
P(B) = 1.26 - 0.89
P(B) = 0.37
Therefore, the probability that a student likes both dogs and cats is 0.37.