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.

- anonymous

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- schrodinger

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- Directrix

Hey, CW.

- anonymous

hey

- anonymous

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

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## More answers

- anonymous

larry must be a popular guy

- anonymous

how ya doin g

- anonymous

doing

- anonymous

these word problems are killing me

- anonymous

it is the same problem with different numbers

- anonymous

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

- anonymous

2 days to complete them

- anonymous

no i am not....

- anonymous

I like to know what i am doing though

- Directrix

Wyatt Earp: Speed is great but accuracy is everything.

- anonymous

exactly

- anonymous

i am always mixing these up as I work through them

- anonymous

I have 12 or 13 problems due in the morning

- anonymous

I feel like a math bum :)

- Directrix

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.

- anonymous

is that the final solution

- anonymous

for both male and female

- anonymous

i know I am very lost haha

- anonymous

it didnt work though

- anonymous

why the zero in both combinations

- anonymous

not quite brain food :) and this one

- anonymous

ha ha i post them wuickly

- anonymous

quickly

- Directrix

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

- anonymous

Correct

- anonymous

So the second part would be similar?

- Directrix

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

- anonymous

no it didnt work

- anonymous

larry wouldnt be 5?

- anonymous

5,1

- Directrix

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.

- Directrix

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

- anonymous

how ya doing

- Directrix

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

- Directrix

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

- anonymous

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

- anonymous

I am wrong no matter what i do on these

- Directrix

Let's find another problem.

- anonymous

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

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