Answer: 
Step-by-step explanation:
Given
Survey shows that 16% of college students have dogs and 38% have HBO subscription
Probability that a random person have both is
![\Rightarrow P_o=0.16\times 0.38\quad [\text{As both events are independent}]\\\Rightarrow P_o=0.0608](https://tex.z-dn.net/?f=%5CRightarrow%20P_o%3D0.16%5Ctimes%200.38%5Cquad%20%5B%5Ctext%7BAs%20both%20events%20are%20independent%7D%5D%5C%5C%5CRightarrow%20P_o%3D0.0608)
The probability that the random person has neither of the two is
Answer:
BC~=AD (GIVEN )
CBA=CDA (altanate angles BC=AD)
C=A (altanate angles (BC=AD)
AAS
brainyliest please
Answer:
20.508%
Step-by-step explanation:
1/2 chance for one free throw.
1/2*4/10=4/20=1/5
Therefore, 20.508%
Answer:
In a certain Algebra 2 class of 30 students, 22 of them play basketball and 18 of them play baseball. There are 3 students who play neither sport. What is the probability that a student chosen randomly from the class plays both basketball and baseball?
I know how to calculate the probability of students play both basketball and baseball which is 1330 because 22+18+3=43 and 43−30 will give you the number of students plays both sports.
But how would you find the probability using the formula P(A∩B)=P(A)×p(B)?
Thank you for all of the help.
That formula only works if events A (play basketball) and B (play baseball) are independent, but they are not in this case, since out of the 18 players that play baseball, 13 play basketball, and hence P(A|B)=1318<2230=P(A) (in other words: one who plays basketball is less likely to play basketball as well in comparison to someone who does not play baseball, i.e. playing baseball and playing basketball are negatively (or inversely) correlated)
So: the two events are not independent, and so that formula doesn't work.
Fortunately, a formula that does work (always!) is:
P(A∪B)=P(A)+P(B)−P(A∩B)
Hence:
P(A∩B)=P(A)+P(B)−P(A∪B)=2230+1830−2730=1330
