binomial and hypergeometric probability distributions:
from a box of 12 flares, four are selected at random. If the box contains four flares that do no work:
a. what is the probability that all four will not work?
b. what is the probability that, at most, two will not work?
c. of the four flares selected, how many would you expect to not work?
probability distribution for discrete random variables:
Jason has designed a game where if you roll a prime number with two six-sided dice, you receive twice the amount of your roll, but if you roll a composite number (not prime), you must must pay the value of the roll.
a. what is the expected value [E(x)] per roll?
b. is this game fair? explain.
Thanks for the help everyone
Hey,
could you please review my solution for the second question I asked?
Jason has designed a game where if you roll a prime number with two six-sided dice, you receive twice the amount of your roll, but if you roll a composite number (not prime), you must must pay the value of the roll.
a. what is the expected value [E(x)] per roll?
2, 3, 5, 7, and 11 are prime, 4, 6, 8, 9, 10, and 12 are composite.
Let X be the variable of the net value of each roll. So if, for instance, you roll a 5, X is 10, but if you roll a 6, X is -6.
is it correct to do it by making this table?
. X ... p(X) .... X*p(X)
-12 ... 1/36 ... -12/36
-10 ... 3/36 ... -30/36
-9 ..... 4/36 ... -36/36
-8 ..... 5/36 ... -40/36
-6 ..... 5/36 ... -30/36
-4 ..... 3/36 ... -12/36
4 ...... 1/36 ... 4/36
6 ...... 2/36 ... 12/36
10 .... 4/36 ... 40/36
14 .... 6/36 ... 84/36
22 .... 2/36 ... 44/36
Add up the third column to get E(X) = 24/36 = 2/3
b. The game is not fair since E(X) is not equal to zero.
I g ave the flare question another shot let's see if I got any of the parts right lol, sorry I'm doing this course all alone so I dont really have anyone to turn to for more help...
a. C(12,4) = 12 x 11 x 10 x 9
4 x 3 x 2 x 1
= 11880 = 495
24
b. 12 – 4 = 8 will work
p(x>2) = 1 – p(x=0) – p(x=1)
p(x>2) = 1 – C(4,0) x C(8,4) – C(4,1) x C(8,3)
C(12,4) C(12,4)
= 1 - .1414141414 - .4525252525
= .4060606061 = 41%
c. 4 x .4060606061
= 1.6 = approximately 2 out of the 4 will not work
Hey Plato, and others,
I am on this unit (that's attached below) and I am really struggling to understand it, in the first example they find the line of best fit and when I tried to find it was different... I am just pretty confused about this whole unit.
It would be great if you could help me out..
Thanks,
Holly
That is in no way what I meant to do, it was not a lazy move. I wrote that I am having trouble with the whole line of best fit equation and such I just wanted to show the work I was working with and that is why I attached the file, it was not meant to be lazy in any way...
Would you like me to post the file in a new thread?