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Out of a box that contains 4 black and 6 white mice

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Out of a box that contains 4 black and 6 white mice  [#permalink]

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New post 20 Sep 2010, 07:37
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Could anyone solve the following with a combinatorial approach instead of probability one?

Out of a box that contains 4 black and 6 white mice, three are randomly chosen. What is the probability that all three will be black?

a) 8/125.
b) 1/30.
c) 2/5.
d) 1/720.
e) 3/10.

Thanks!
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Re: Out of a box that contains 4 black and 6 white mice  [#permalink]

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New post 20 Sep 2010, 07:48
rraggio wrote:
Could anyone solve the following with a combinatorial approach instead of probability one?

Out of a box that contains 4 black and 6 white mice, three are randomly chosen. What is the probability that all three will be black?

a) 8/125.
b) 1/30.
c) 2/5.
d) 1/720.
e) 3/10.

Thanks!


not sure what you mean by "combinatorial approach", but this is how you can solve it using counting :

Number of ways to choose 3 black mice = \(C^4_3=4\) (Picking any three of the four)

Number of ways to choose any 3 mice = \(C^{10}_3=120\) (Picking any three of the ten)

So probability that all three mice are black is 4/120 or 1/30
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Re: Out of a box that contains 4 black and 6 white mice  [#permalink]

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New post 20 Sep 2010, 08:16
Without bringing formula into play I use below method

_ _ _
We have to fill 3 places.


Number of ways we can pick any mice is
10 * 9 * 8
__ __ __

Number of ways we can have all 3 black mice
4 * 3 * 2
__ __ __
So 1/30

PS - I know this is another way of C(4,3) and C(10,3) :)...
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Re: Out of a box that contains 4 black and 6 white mice  [#permalink]

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New post 21 Sep 2010, 02:26
Thanks, you got it.

Since I'm very weak with combinations and probability, I'm trying to solve it in the following ways.
Three of four ways (1,2,4) work, one doesn't (3).
Could anybody say what's wrong in my third approach?
Thanks!

1) probability approach:
picking first black mice = \(4/10\) = \(2/5\)
picking second black mice = \(3/9\) = \(1/3\)
picking third black mice = \(2/8\) = \(1/4\)

picking three black mice = \(2/5*1/3*1/4=1/30\)

2) combinatorial approach:
Quote:
Number of ways to choose 3 black mice = \(C(4,3)\) (Picking any three of the four)
Number of ways to choose any 3 mice = \(C(10,3)\) (Picking any three of the ten)
So probability that all three mice are black is \(4/120\) or \(1/30\)


3) reversal probability approach:
The probability of picking three black mices equals to 1 minus the probability of picking three white mices plus the probability of picking two white mices and one black plus the probability of picking one white mices and two black.
That is:
P(b,b,b) = 1 - [P(w,w,w) + P(w,w,b) + P(w,b,b)]
P(w,w,w) = \(6/10*5/9*4/8=1/6\)
P(w,w,b) = \(6/10*5/9*4/8=1/6\)
P(w,b,b) = \(6/10*4/9*3/8=1/10\)

so:
P(b,b,b) = \(1- (1/6+1/6+1/10) = 1 - 13/30 = 17/30\)

4) reversal combinatorial approach:
Same approach of the above one.

Number of ways to choose any 3 mice = \(C(10,3)\) = 120 (Picking any three of the ten)
Number of ways to choose one mice white and two mices black = \(C(6,1)\)*\(C(4,2)\) = 36
Number of ways to choose two white mices and one mice white = \(C(6,2)\)*\(C(4,1)\) = 60
Number of ways to choose three white mices = \(C(6,3)\) = 20

So probability that all three mice are black is:
\(1 - [C(6,1)*C(4,2)+C(6,2)*C(4,1)+C(6,3)]/C(10,3) = 1 - (36+60+20)/120 = 4/120 = 1/30\)
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Re: Out of a box that contains 4 black and 6 white mice  [#permalink]

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New post 21 Sep 2010, 02:34
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rraggio wrote:
Thanks, you got it.

Since I'm very weak with combinations and probability, I'm trying to solve it in the following ways.
Three of four ways (1,2,4) work, one doesn't (3).
Could anybody say what's wrong in my third approach?
Thanks!
3) reversal probability approach:
The probability of picking three black mices equals to 1 minus the probability of picking three white mices plus the probability of picking two white mices and one black plus the probability of picking one white mices and two black.
That is:
P(b,b,b) = 1 - [P(w,w,w) + P(w,w,b) + P(w,b,b)]
P(w,w,w) = \(6/10*5/9*4/8=1/6\)
P(w,w,b) = \(6/10*5/9*4/8=1/6\)
P(w,b,b) = \(6/10*4/9*3/8=1/10\)

so:
P(b,b,b) = \(1- (1/6+1/6+1/10) = 1 - 13/30 = 17/30\)


If you calculate probability in this manner, the order of drawing is inherently important. So you need to rule out the probability of the following as well :
P(w,b,w); P(b,w,w); P(b,w,b); P(b,b,w)
These will turn out to be 1/6,1/6,1/10,1/10
Subtracting this, once again you will get 1/30
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Re: Out of a box that contains 4 black and 6 white mice  [#permalink]

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New post 07 Jan 2018, 08:18
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rraggio wrote:
Out of a box that contains 4 black and 6 white mice, three are randomly chosen. What is the probability that all three will be black?

a) 8/125.
b) 1/30.
c) 2/5.
d) 1/720.
e) 3/10.

Thanks!


P(all 3 selected mice are black) = P(1st mouse is black AND 2nd mouse is black AND 3rd mouse is black)
= P(1st mouse is black) x P(2nd mouse is black) x P(3rd mouse is black)
= 4/10 x 3/9 x 2/8
= 1/30
= B

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Re: Out of a box that contains 4 black and 6 white mice  [#permalink]

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New post 07 Jan 2018, 10:16
Top Contributor
rraggio wrote:
Could anyone solve the following with a combinatorial approach instead of probability one?

Out of a box that contains 4 black and 6 white mice, three are randomly chosen. What is the probability that all three will be black?

a) 8/125.
b) 1/30.
c) 2/5.
d) 1/720.
e) 3/10.

Thanks!


Here's the counting approach:

P(all 3 mice are black) = (# of ways to select 3 black mice)/(TOTAL # of ways to select ANY 3 mice)

# of ways to select 3 black mice
Since the order in which we select the mice does not matter, we can use COMBINATIONS
We can select 3 black mice from 4 black mice in 4C3 ways (= 4 ways)

TOTAL # of ways to select ANY 3 mice
We can select 3 mice from all 10 mice in 10C3 ways (= 120 ways)

ASIDE: If anyone is interested, we have a video on calculating combinations in your head (below)

So, P(all 3 mice are black) = 4/120
= 1/30
= B

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Re: Out of a box that contains 4 black and 6 white mice  [#permalink]

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New post 07 Jan 2018, 11:33
1
selection of three mice: 4C3

Total selection of mice: 10C3

Prob : 4C3/10C3 = 1/30
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Re: Out of a box that contains 4 black and 6 white mice  [#permalink]

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New post 07 Jan 2018, 18:22
rraggio wrote:
Could anyone solve the following with a combinatorial approach instead of probability one?

Out of a box that contains 4 black and 6 white mice, three are randomly chosen. What is the probability that all three will be black?

a) 8/125.
b) 1/30.
c) 2/5.
d) 1/720.
e) 3/10.

Thanks!

---------
A box contains 4 black and 6 white mice.
Using probability approach to determine the probability that all three will be black?

First selection: 4/10 (4B, 6W)
Second selection: 3/9 (3B, 6W)
Third selection: 2/8 (2B, 6W)

Combined = (4/10)*(3/9)*(2/8) = 1/30
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Re: Out of a box that contains 4 black and 6 white mice  [#permalink]

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New post 10 Mar 2018, 11:29
Hi All,

You can approach this question in a couple of different ways. If you're going to treat it as a straight-forward probability question, then here's how you would do the math:

4 Black
6 White
10 Total

It's important to remember that once you choose a mouse, there will be one FEWER mouse for the next pick. The probability of choosing 3 black mice is…

(4/10)(3/9)(2/8) = 1/30

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Re: Out of a box that contains 4 black and 6 white mice &nbs [#permalink] 10 Mar 2018, 11:29
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