A cake shop bakes a variety of brownies. The top-selling brownies are ones with toppings of chocolate chip, walnuts, or both. A
customer enters the store. The probability that the customer will pick both toppings is 0.4. What is the probability that they will pick neither the chocolate chip nor the walnut toppings?
2 answers:
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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
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,
Answer:
Step-by-step explanation:
Hello
let's say that the price of the book was x
the price after a 20% discount is x - 20%*x = x*(1-20%)=x*(1-.20)=0.8*x
and this is $72 so we can write that
0.8*x=72
and then divide by 0.8 both parts
x = 72/0.8=90
So the list price value of the book is $90
and we can verify as 90 - 20%*90 = 90 - 18 = 72
Hope this helps