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# If you flip a fair coin 4 times, what is the probability that you will get exactly 2 tails?

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If you flip a fair coin 4 times, what is the probability that you will get exactly 2 tails?

Guest May 22, 2014

#7
+18827
+5

Hi Melody,

here is the code:

\documentclass[7pt,a4paper,landscape]{article}
\usepackage[utf8]{inputenc}
\usepackage{pgfplots}
\usepackage{qtree}

\begin{document}
\Tree [
[
[
[[ F\\F\\F\\F ].($\frac{1}{2}$)\\F [ F\\F\\F\\B ].($\frac{1}{2}$)\\B ].($\frac{1}{2}$)\\F [[ F\\F\\B\\F ].($\frac{1}{2}$)\\F [ \fbox{  \textcolor[rgb]{1,0,0}{FFBB}} ].($\frac{1}{2}$)\\B ].($\frac{1}{2}$)\\B
].($\frac{1}{2}$)\\F
[
[[ F\\B\\F\\F ].($\frac{1}{2}$)\\F [ \fbox{ \textcolor[rgb]{1,0,0}{FBFB}} ].($\frac{1}{2}$)\\B ].($\frac{1}{2}$)\\F [[ \fbox{ \textcolor[rgb]{1,0,0}{FBBF}} ].($\frac{1}{2}$)\\F [ F\\B\\B\\B ].($\frac{1}{2}$)\\B ].($\frac{1}{2}$)\\B
].($\frac{1}{2}$)\\B
].($\frac{1}{2}$)\\F
[
[
[[ B\\F\\F\\F ].($\frac{1}{2}$)\\F [  \fbox{ \textcolor[rgb]{1,0,0}{BFFB}} ].($\frac{1}{2}$)\\B ].($\frac{1}{2}$)\\F [[ \fbox{ \textcolor[rgb]{1,0,0}{BFBF}} ].($\frac{1}{2}$)\\F [ B\\F\\B\\B ].($\frac{1}{2}$)\\B ].($\frac{1}{2}$)\\B
].($\frac{1}{2}$)\\F
[
[[\fbox{ \textcolor[rgb]{1,0,0}{BBFF}} ].($\frac{1}{2}$)\\F [ B\\B\\F\\B ].($\frac{1}{2}$)\\B ].($\frac{1}{2}$)\\F [[ B\\B\\B\\F ].($\frac{1}{2}$)\\F [ B\\B\\B\\B ].($\frac{1}{2}$)\\B ].($\frac{1}{2}$)\\B
].($\frac{1}{2}$)\\B
].($\frac{1}{2}$)\\B
].Start\\(F=Front\\B=Back)
\end{document}

heureka

heureka  May 26, 2014
Sort:

#1
+80875
+5

6/16

To see this notice that a tail could occur on the first flip and then on any of the next three flips. So there are three ways that can happen.

Or, it could occur on the second flip and then on any of the two remaining flips. That's two more ways.

Or, it could occur on the third and on the 4th flips.

So, 3 + 2 + 1 = 6

And there are 2 outcomes possible on each flip, so the total number of outcomes is just 2*2*2*2 = 16

Thus,   6/16

CPhill  May 22, 2014
#2
+18827
+5

$$\dbinom {4} {2} * \dfrac{1}{2} * \dfrac{1}{2}* \dfrac{1}{2}* \dfrac{1}{2} = \dbinom {4} {2} * \left(\dfrac{1}{2}\right)^4=6*\dfrac{1}{16}=\dfrac{6}{16}=\dfrac{3}{8}=0.375$$

The probability is 37.5 %

heureka  May 22, 2014
#3
+18827
+5

6 times you have exactly two tails: FFBB, FBFB, FBBF, BFFB, BFBF, BBFF

And the Probability is  $$6*\left(\dfrac{1}{2}\times\dfrac{1}{2}\times\dfrac{1}{2}\times\dfrac{1}{2} \right)$$

heureka  May 23, 2014
#4
+91436
0

Hi Heureka,

Your tree is very impressive!

What program did you use to draw it?  Can you draw one from left to right as well?

Melody  May 23, 2014
#5
+18827
+5

Hi Melody,

I use Latex \usepackage{qtree}

Sorry, i can't answer your second question.

My program lines:

Bye

heureka

heureka  May 26, 2014
#6
+91436
0

Thank you very much Heureka!

Could you please copy the actual code in here so I can copy it into Texmaker and play with it.

Besides being lazy my eyes are not that good and I am having a hard time reading it off the screen.

Also, I have reference your last post in the Latex thread in the Sticky Notes.

Thank you.

Melody  May 26, 2014
#7
+18827
+5

Hi Melody,

here is the code:

\documentclass[7pt,a4paper,landscape]{article}
\usepackage[utf8]{inputenc}
\usepackage{pgfplots}
\usepackage{qtree}

\begin{document}
\Tree [
[
[
[[ F\\F\\F\\F ].($\frac{1}{2}$)\\F [ F\\F\\F\\B ].($\frac{1}{2}$)\\B ].($\frac{1}{2}$)\\F [[ F\\F\\B\\F ].($\frac{1}{2}$)\\F [ \fbox{  \textcolor[rgb]{1,0,0}{FFBB}} ].($\frac{1}{2}$)\\B ].($\frac{1}{2}$)\\B
].($\frac{1}{2}$)\\F
[
[[ F\\B\\F\\F ].($\frac{1}{2}$)\\F [ \fbox{ \textcolor[rgb]{1,0,0}{FBFB}} ].($\frac{1}{2}$)\\B ].($\frac{1}{2}$)\\F [[ \fbox{ \textcolor[rgb]{1,0,0}{FBBF}} ].($\frac{1}{2}$)\\F [ F\\B\\B\\B ].($\frac{1}{2}$)\\B ].($\frac{1}{2}$)\\B
].($\frac{1}{2}$)\\B
].($\frac{1}{2}$)\\F
[
[
[[ B\\F\\F\\F ].($\frac{1}{2}$)\\F [  \fbox{ \textcolor[rgb]{1,0,0}{BFFB}} ].($\frac{1}{2}$)\\B ].($\frac{1}{2}$)\\F [[ \fbox{ \textcolor[rgb]{1,0,0}{BFBF}} ].($\frac{1}{2}$)\\F [ B\\F\\B\\B ].($\frac{1}{2}$)\\B ].($\frac{1}{2}$)\\B
].($\frac{1}{2}$)\\F
[
[[\fbox{ \textcolor[rgb]{1,0,0}{BBFF}} ].($\frac{1}{2}$)\\F [ B\\B\\F\\B ].($\frac{1}{2}$)\\B ].($\frac{1}{2}$)\\F [[ B\\B\\B\\F ].($\frac{1}{2}$)\\F [ B\\B\\B\\B ].($\frac{1}{2}$)\\B ].($\frac{1}{2}$)\\B
].($\frac{1}{2}$)\\B
].($\frac{1}{2}$)\\B
].Start\\(F=Front\\B=Back)
\end{document}

heureka

heureka  May 26, 2014
#8
+91436
0

Thank you Heureka

Melody  May 26, 2014

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