Please Help me with this Probability problems???

Question

I have been try to solve this 2 problems. I need someone to help me. The first one is: An office has 12 female workers and 6 male workers. If the boss randomly picks workers for a special job, find the probability that 2 males and 2 females will be pick. So far I come up with 7/9 as the answer for that one.

The second one is: two fair dice are rolled. What is the probability the sum is 7, if we are given that each die comes up less that 5. I was thinking that the anwser was three times.
Can someone please help me if I am wrong. Thank you very much.

Answer 1

1) (12/18)*(11/17) *(6/16)*(5/15)

2) If you know both dice are less than 5, then you can only get seven if 1= 4 and the other =3. Since there are 2 ways you can have that combination with 2 dice, and 36 possibilities = 2/36

Answer 2

If each die is less than 5, you only have 1 through 4 to work with. That means there are 4 * 4 ways they can be, and of those, only a three on one and a 4 on the other add to 7. Since the first or the second could be the three and the other one the 4, you have a 2 out of 16 or 1/8 chance.

Answer 3

I know that your answer for the first one is wrong, but I can't remember how to do that. For the second one, figure out all of the combinations that would work, like 3-4 and 4-3? Then you have two options for the first die (3 or 4) and after that, only one option for the second die for it to add to 7 and not be over 5, so this equation would be (2/6) * (1/6) I think! But it has been a while!!

Answer 4

1. Usually the order people are picked in problems like this doesn't matter, so combinations are what you need. I'll write
C(n,r) for the number of ways to choose r out of n things.

The denominator of the probability fraction is C(18,4), the number of ways to choose four employees. For the numerator, two of each gender can be chosen C(12,2)*C(6,2) ways. That gives (66*15)/3060 ~ .3235

2. Since you know each die is less than 5 (i.e., 1, 2, 3, or 4), the sample space is only 16 rolls, not the normal 36. There are just two ways to get 7 then, (3,4) and (4,3). So the probability is 2 / 16 = .125

Note: This is all the same as Blah31 (I was working on this when it came up), except I think she has an error in computing C(18,4). After cancelling out the 14!, It's (18*17*16*15)/(4*3*2*1) = 73440/24 = 3060.

Answer 5

The first one is 1/3

Answer 6

The number of ways that you can select four workers from a total of 18 is 18!/(4!14!) = 3060 ways. Of those ways, the number of ways that two males and two females are selected is {12!/(2!10!)} * {6!/(2!4!)} = 990 ways, so the probability is

990/3060 = 11/34

The 990 comes from the General Counting Principle; we should multiply the number of ways to pick the 2 females and 2 males. There are 12!/(2!10!) ways to pick the two females and 6!/(2!4!) ways to pick the two males.

The possible ways that the two dice can come up is

1,1 1,2 1,3 1,4
2,1 2,2 2,3 2,4
3,1 3,2 3,3 3,4
4,1 4,2 4,3 4,4

since both dice have to be less than 5. All of these are equally likely. Of these 16 ways, only two sum to 7 (namely 3,4 and 4,3), so the probability is 2/16 or 1/8.

edit: Thanks Brashion. Stupid calculator (at least that's what I blame it on).

Answer 7

i choose the ten undesirable--SO i assumed this one outONEOUT AND MY answer IS.........27% threat OF GETTIN THE ALL HEARTS IN sequence FROM AN ACE to ten. it is HOW I ARRIVED at my determination, ok???? pay attention nicely reason i only madeit to point 3 after too long of a time and now i'm attempting point 4. if i had fifty two enjoying cards that are in a deck of enjoying cards and u dealt me 5 like u stated u did--that ought to go away me with 40 seven enjoying cards. minus a 10 could equivalent 37p.c.and the possibilities on that are even worse-r than that so upload yet another or sub-song yet another 10 for the possibilities bein against me and that i only ended up with a 27% and am i ceremony or rong??? i'm a kia