SOLUTION: 1. if four coins are flipped at once, what is the probability that exactly 3 tails will come up? 2.What is 6x+6y/x-y ___________ 18/5x-5y 3. What is th

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Question 145031: 1. if four coins are flipped at once, what is the probability that exactly 3 tails will come up?
2.What is 6x+6y/x-y
____________
18/5x-5y
3. What is the probability of rolling two number cubes and getting the sum that is odd or a multiple of 3?
Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
1. if four coins are flipped at once, what is the probability that exactly 3 tails will come up?

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,HTTT
 THHH,THHT,THTH,THTT
 TTHH,TTHT,TTTH,TTTT}

There are 4 out of 16, so the probabliity is 4%2F16 or 1%2F4

---------------

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%28n%2Cx%2Cp%29+=+%28n%21%2F%28r%21%28n-r%29%21%29%29%28p%29%5Er%2A%281-p%29%5E%28n-r%29

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%2F2, is



P%284%2C3%2C1%2F2%29+=+%284%21%2F%283%211%21%29%29%281%2F8%29%2A%281%2F2%29%5E%281%29





P%284%2C3%2C1%2F2%29+=+4%281%2F8%29%2A%281%2F2%29+

P%284%2C3%2C1%2F2%29+=+4%281%2F16%29+

P%284%2C3%2C1%2F2%29+=+4%2F16

P%284%2C3%2C1%2F2%29+=+1%2F4

---------------------------

2.What is %28%286x%2B6y%29%2F%28x-y%29%29%2F%2818%2F%285x-5y%29%29

Write as a division:

%28%286x%2B6y%29%2F%28x-y%29%29÷%2818%2F%285x-5y%29%29

Invert the second fraction and change the division to multiplication:

%28%286x%2B6y%29%2F%28x-y%29%29%2A%28%285x-5y%29%2F18%29

Factor 6 out of 6x%2B6y%29 and 5 out of 5x-5y  

%28%286%28x%2By%29%29%2F%28x-y%29%29%2A%28%285%28x-y%29%29%2F18%29

Cancel the 6 into the 18 getting 3, and cancel the %28x-y%29s:

 
            3

%28%28x%2By%295%29%2F3

%285%28x%2By%29%29%2F3

You can either leave it like that or

distribute the top out and leave it like this:

%285x%2B5y%29%2F3

Or you can write %285%28x%2By%29%29%2F3 as 5%2F3%28x%2By%29

3. What is the probability of rolling two number cubes and getting the sum that is odd or a multiple of 3?

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) (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)

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%2F36 which reduces to 2%2F3.

Edwin