This math question

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<p>Of 5 employees, 3 are to be assigned an office and 2 are to be assigned a cubicle. If 3 of the employees are men and 2 are women, and if those assigned an office are to be chosen at random, what is the probability that the offices will be assigned to 2 of the men and 1 of the women?</p>

<p>1/3
2/5
1/2
3/5
2/3</p>

<p>The explanation given by CB is pretty complicated, so anyone have a more doable approach to this problem? Thanks.</p>

2

<p>I didn't get that 1 either.....missed it.</p>

3

<p>ugh! spent 5 mins, i got 1/5,which is not there lol...
then i got 1/3, wats the correct answer??
the way i did it is 2 men out of 3 men: 2/3
1 woman out of 2 women: 1/2
2/3 x 1/2 = 1/3??</p>

4

<p>(3 C 2 (Men) * 2 C 1 (Women)) / (5 C 3) (total possibilities) </p>

<p>u get 6/10 which is 3/5...I used probability stuff i learned in pre-cal.</p>

5

<p>i think i agree with ^</p>

6

<p>I think the answer should be 1/3.
Out of the 5 employees 3 are men and two are women and three are to be assigned an office.
So, after assigning the three employees the possible outcomes are:
1) All 3 men
2) 1 man and two women
3) Two men and one woman</p>

<p>So, the required outcome is the third one.
and, No. of total outcomes = 3
Hence, the probability 1/3.</p>

7

<p>Yeah the answer is 3/5. SAS- Howd you know to do that?</p>

8

<p>I had to do it by grunt work:</p>

<p>m1 m2 m3 F1 F2
m1 m2 f1 M3 F2
m1 m2 f2 M3 F1
m1 m3 f1 M2 F2
m1 m3 f2 M2 F1
m1 f1 f2 M2 M3
m2 m3 f1 M1 F2
m2 m3 f2 M1 F1
m2 f1 f2 M1 M3
m3 f1 f3 M1 M2</p>

<p>10 combos w/ 6 having 2 male and 1 female therefore 6/10 = 3/5</p>

9

<p>3c2 * 2c1 / 5c3</p>

<p>= 3/5</p>

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<p>How do you know to do that?</p>

11

<p>^ You learn more about probabilty in Pre-Calculus.</p>

12

<p>I learned probability in Alg II, but since that was like over two years ago, I don't remember it.</p>

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<p>Xcellerator- I took Pre Calc honors. And got an A. But never really understood how to use combination and stuff.</p>

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<p>3c2 means ur selecting 2 frm 3 men and 2c1 means selecting 1 woman frm 2. Multiplyin the two gives no. Of ways for making the required selection as per d question. And total ways of selecting 3 frm 5 ppl is 5c3.</p>

15

<p>Combinations =/ i hate those too xD</p>

16

<p>gah.. this is probability from algebra I... I HATED probability.</p>

17

<p>Damn, man, I took PreCalculus Honors with Trig, and I never learned any of this 5c3 stuff. i dont go to a ■■■■■■ school either, i go to a school which was ranked 89th best public high school or something in 2007 (Oakton High).</p>

18

<p>weird question…</p>

<p>

thats the answer I would have come up with.</p>

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<p>@ Goez , </p>

<p>yes but you must realize there is only 1 way to do your “option 1” 3 ways to do your “option 2” and 6 ways to do “option 3” </p>

<p>6/10 ways are to have 2 men and one woman (which reduces to 3/5) </p>

<p>however I am also not to sure about combinations and I took honors precalc and trig as well, I simply just did this problem in my head took me 1min></p>

20

<p>Of 5 employees, 3 are to be assigned an office and 2 are to be assigned a cubicle. If 3 of the employees are men and 2 are women, and if those assigned an office are to be chosen at random, what is the probability that the offices will be assigned to 2 of the men and 1 of the women?</p>

<p>1/3
2/5
1/2
3/5
2/3</p>

<p>3C2* 2C1 / 5C3
= 3<em>2/ (5</em>4/2)
= 3/5</p>