Question 145031
1. if four coins are flipped at once, what is the probability that exactly 3 tails will come up?
<pre><font size = 4 color = "indigo"><b>
Two ways to do the problem, depending on what you are studying.
List the sample space:

{HHHH,HHHT,HHTH,HHTT
 HTHH,HTHT,HTTH,HTTT
 THHH,THHT,THTH,THTT
 TTHH,TTHT,TTTH,TTTT}

Now I will color the ones red which have exactly 3 tails:

{HHHH,HHHT,HHTH,HHTT
 HTHH,HTHT,HTTH,<font color="red">HTTT</font>
 THHH,THHT,THTH,<font color="red">THTT</font>
 TTHH,<font color="red">TTHT</font>,<font color="red">TTTH</font>,TTTT}

There are 4 out of 16, so the probabliity is {{{4/16}}} or {{{1/4}}}

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Second way.  Only use this is you are studying binomial probability.
Otherwise ignore it. 

The formula for the probability of getting exactly x successes out of n
independent trials, when the probability of getting one success in one 
trial is p, is

{{{P(n,x,p) = (n!/(r!(n-r)!))(p)^r*(1-p)^(n-r)}}}

The formula for the probability of getting exactly 3 successes out of 4
independent trials, when the probability of getting one success in one 
trial is {{{1/2}}}, is

{{{P(4,3,1/2) = (4!/(3!(4-3)!))(1/2)^3*(1-1/2)^(4-3)}}}

{{{P(4,3,1/2) = (4!/(3!1!))(1/8)*(1/2)^(1)}}}

{{{P(4,3,1/2) = ((4*3*2*1)/(3*2*1*1))(1/8)*(1/2)}}}

{{{P(4,3,1/2) = ((4*cross(3*2*1))/(cross(3*2*1)*1))(1/8)*(1/2)}}}

{{{P(4,3,1/2) = 4(1/8)*(1/2) }}}

{{{P(4,3,1/2) = 4(1/16) }}}

{{{P(4,3,1/2) = 4/16}}}

{{{P(4,3,1/2) = 1/4}}}

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</pre></font></b> 
2.What is {{{((6x+6y)/(x-y))/(18/(5x-5y))}}}
<pre><font size = 4 color = "indigo"><b>
Write as a division:

{{{((6x+6y)/(x-y))}}}÷{{{(18/(5x-5y))}}}

Invert the second fraction and change the division to multiplication:

{{{((6x+6y)/(x-y))*((5x-5y)/18)}}}

Factor {{{6}}} out of {{{6x+6y)}}} and {{{5}}} out of {{{5x-5y}}}  

{{{((6(x+y))/(x-y))*((5(x-y))/18)}}}

Cancel the {{{6}}} into the {{{18}}} getting 3, and cancel the {{{(x-y)}}}s:

{{{((cross(6)(x+y))/cross((x-y)))*((5cross((x-y)))/cross(18))}}} 
            {{{3}}}

{{{((x+y)5)/3}}}

{{{(5(x+y))/3}}}

You can either leave it like that or

distribute the top out and leave it like this:

{{{(5x+5y)/3}}}

Or you can write {{{(5(x+y))/3}}} as {{{5/3}}}{{{(x+y)}}}
</pre></font></b>
3. What is the probability of rolling two number cubes and getting the sum that is odd or a multiple of 3?
<pre><font size = 4 color = "indigo"><b>
Here are all the ways a pair of fair dice (number cubes) can be rolled:

(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)

(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)

(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)

(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)

(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)

(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)

That is the sample space.  There are 36
outcomes in that sample space.

Now I will go through and color the ones
red which are either an odd number or a 
multiple of 3)

(1,1) <font color="red">(1,2)</font> (1,3) <font color="red">(1,4)</font> <font color="red">(1,5)</font> <font color="red">(1,6)</font>

<font color="red">(2,1)</font> (2,2) <font color="red">(2,3)</font> <font color="red">(2,4)</font> <font color="red">(2,5)</font> (2,6)

(3,1) <font color="red">(3,2)</font> <font color="red">(3,3)</font> <font color="red">(3,4)</font> (3,5) <font color="red">(3,6)</font>

<font color="red">(4,1)</font> <font color="red">(4,2)</font> <font color="red">(4,3)</font> (4,4) <font color="red">(4,5)</font> (4,6)

<font color="red">(5,1)</font> <font color="red">(5,2)</font> (5,3) <font color="red">(5,4)</font> (5,5) <font color="red">(5,6)</font>

<font color="red">(6,1)</font> (6,2) <font color="red">(6,3)</font> (6,4) <font color="red">(6,5)</font> <font color="red">(6,6)</font>

There are 24 throws out of the 36 that are either odd or are 
a multiple of 3.  

24 successful ways out of 36 possible ways gives a probability
of {{{24/36}}} which reduces to {{{2/3}}}.

Edwin</pre>