Question

7 females, and 5 males including Larry.
There are 4 tasks to be assigned. Note that assigning the same
people different tasks constitutes a different assignment.
(1) Find the probability that both males and females are given a task.
(2) Find the probability that Larry and at least one female
are given tasks.

Answer 1

Hey, CW.

Answer 2

hey

Answer 3

wow check this out
http://www.jiskha.com/display.cgi?id=1236051521

Answer 4

larry must be a popular guy

Answer 5

how ya doin g

Answer 6

doing

Answer 7

these word problems are killing me

Answer 8

it is the same problem with different numbers

Answer 9

I had to take an exam today. And this homework set is due on tomorrow at 730 am. :)

Answer 10

2 days to complete them

Answer 11

no i am not....

Answer 12

I like to know what i am doing though

Answer 13

Wyatt Earp: Speed is great but accuracy is everything.

Answer 14

exactly

Answer 15

i am always mixing these up as I work through them

Answer 16

I have 12 or 13 problems due in the morning

Answer 17

I feel like a math bum :)

Answer 18

Don't feel. Just work. Post the other 12 or 13 problems individually on this site. You do that while I work on this problem. I may have to stop and eat some brain food. :} You need to focus. Nobody here is a bum.

Answer 19

is that the final solution

Answer 20

for both male and female

Answer 21

i know I am very lost haha

Answer 22

it didnt work though

Answer 23

why the zero in both combinations

Answer 24

not quite brain food :) and this one

Answer 25

ha ha i post them wuickly

Answer 26

quickly

Answer 27

P(both genders) = 1 - { [ C(7,0) C(5,4) - C(7,4) C(5,0)] / C(12,4) } = 1 - 40/495 = 455/495 = 91/99 = .919 approx

Answer 28

Correct

Answer 29

So the second part would be similar?

Answer 30

P(Larry and at least 1 F)

P(at least 1F) = 1 - P(0F) = 1 -{ [C(7,0)C(5,4)] } / C(12,4) = 5/495 = 490/495 = 98/99 = .99 ish.

Ways to choose Larry = 1 C(1,1) = 1

P(Larry and at least 1 F) = 1(98/99) = 98/99 = .99 approx

Answer 31

no it didnt work

Answer 32

larry wouldnt be 5?

Answer 33

5,1

Answer 34

But, we want Larry and none of the other males. So P(Larry) = C(1,1) C(4,0) = 1. That would be Larry and none of the other males.

Answer 35

I'm going to cases for this. Slow but may make more sense.

Answer 36

how ya doing

Answer 37

Larry and at least 1F
--------
Cases
{Lar, 1F, 2 other M}; {Lar, 2F, 1 other M}; and {Lar, 3F, 0 other M}

Answer 38

Ways to choose Lar, 1F, 2 other M} = 1 C(7,1) C(4,2) = 42
Ways to choose {Lar, 2F, 1 other M}; = 1 C(7,2) C(4,1)= 84
Ways to choose {Lar, 3F, 0 other M} = 1 C(7,3) C(4,0) = 35
Ways to choose 4 from 12 is C(12,4) = 495

P(L and at least 1F) = [42 + 84 +35] / 495 = 161/495 =.325 approx

Answer 39

CORRECT! You are a genius....
I have posted some more.
I haven't wanted to post to many at a time and flood the screen

Answer 40

I am wrong no matter what i do on these

Answer 41

Let's find another problem.

Answer 42

I used a lot of the work i learned from hear that helped me on the test in some of these.