Answer: 0.05
Step-by-step explanation:
Let M = Event of getting an A in Marketing class.
S = Event of getting an A in Spanish class,
i.e. P(M) = 0.80 , P(S) = 0.60 and P(M∩S)=0.45
Required probability = P(neither M nor S)
= P(M'∩S')
= P(M∪S)' [∵P(A'∩B')=P(A∪B)']
=1- P(M∪S) [∵P(A')=1-P(A)]
= 1- (P(M)+P(S)- P(M∩S)) [∵P(A∪B)=P(A)+P(B)-P(A∩B)]
= 1- (0.80+0.60-0.45)
= 1- 0.95
= 0.05
hence, the probability that Helen does not get an A in either class= 0.05
To solve for the system of equations, I will write the equation down as I rewrite the written form.
a number, n, (n) is added to 15 less than 3 times itself (+3n -15). The result is (=) 101. (101)
Now let's write only what's in the parenthesis.
n + 3n -15 = 101.
The correctly written form in your answers is:
3n - 15 + n = 101, your first answer.
Answer:
The expected value of Jordan gains is -1 dollar.
Step-by-step explanation:
Consider the following random variables. X := #of shots that Jordan makes. Then, we can can define the random variable Y of the earnings of Jordan in a game as follows
Y = 5 if X=2 (since he gets 10, but invested 5), Y=0 if X=1(since he gets the 5 back) and Y=-5 if X=0(since he doesn't get the money back). Then, in this case, we can define the probability as follows.
P(Y=5) = P(X=2), P(Y=0) = P(X=1), P(Y=-5)= P(X=0).
By definition, the expected value of Y is given by
. By the previous analysis, we have that
![E[Y] = 5\cdot P(X=2)-5P(X=0)](https://tex.z-dn.net/?f=E%5BY%5D%20%3D%205%5Ccdot%20P%28X%3D2%29-5P%28X%3D0%29)
We only need to calculate the probabilities for X. In this case, we can consider each shot independt from each other. Then, we can consider X to be distributed as a binomial random variable with n=2 trials and p=0.4 of success (since he has a 40% chance of winning).
Then, by definition

where 
Then,


Then,
![E[Y] = 5\cdot 0.16-5\cdot 0.36 = -1](https://tex.z-dn.net/?f=E%5BY%5D%20%3D%205%5Ccdot%200.16-5%5Ccdot%200.36%20%3D%20-1)
33.68
27.4
x
--------
922.832