But CPhill if we will play the game by your answer it is mean that we will get to the profit
let s say that i woned the first one and loosed the second one ,When i won i will get 1$ and when i loosed i will get -1$
1$-1$=0
i think:
\(1, 2,3,4,5,6\)
there is 3 even numbers out of 6 so the probability to win is 1/2
if you won you recieve 2-1=1$
if you will loose you recieve -1$
0.5*1-0.5*1= 0 $
actually there is easier way to think about it when we devide 10 to 5 is 2 if we will change the question how many times does 10 include 5 the answer is still 2 .So how many times does 1 include 0 infinetly many times . But there is problem with the negative numbers .(i know that i am wrong but i wanna to argue and to make it clear for myself :D)
none because there is no more Isabella
2.
1+6=7
2+5=7
3+4=7
------------------------------
4+3=7
5+2=7
6+1=7
there is 36 posibilities:
\(\frac6{36}=\frac16\)
Answer: (C)
\(C=7\pi\\ C=22\ (aprox.)\)
yes I understood :) , but the question is still wrong we didn t know how many black cards are there :D
yes it is the most suitable choice
yes if you will put it again into deck
And the probability of taking all the 4 black is:
(1/2)*(25/51)*(12/25)*(23/49) = 0.0552220888355342 (6% chance)
