A tree diagram represents the outcomes from a multistep experiment.
A sample tree diagram is shown below.
Repeatedly click the button to get many more (randomlygenerated) samples!
A multistep experiment has more than one step.
For example, here are some twostep experiments:
The branches emanating from any point on a tree diagram must have probabilities
that sum to 1 .
In the sample tree diagram above (initial configuration), we have:
To find the probability of any path,
multiply the probabilities on the corresponding branches.
In the sample tree diagram above (initial configuration),
there are six paths (going from top to bottom):
TWO FLIPS OF A FAIR COIN
Consider the twostep experiment ‘two flips of a fair coin’. Note that the sample space for this experiment is:
$\,S = \{\text{HH},\text{HT},\text{TH},\text{TT}\,\}\,$
The probability tree diagram is shown at right.
$\cssId{s41}{P(\text{both heads})}
\cssId{s42}{= P(\text{HH})}
\cssId{s43}{= (0.5)(0.5)}
\cssId{s44}{= 0.25}$
$\cssId{s45}{P(\text{both tails})}
\cssId{s46}{= P(\text{TT})}
\cssId{s47}{= (0.5)(0.5)}
\cssId{s48}{= 0.25}$
$\cssId{s49}{P(\text{no heads})}
\cssId{s50}{= P(\text{TT})}
\cssId{s51}{= 0.25}$
$P(\text{at least one head})$
$= P(\text{HH or HT or TH})$ $= P(\text{HH}) + P(\text{HT}) + P(\text{TH})$ $= 0.25 + 0.25 + 0.25 = 0.75$
(alternate method)
$P(\text{at least one head})$ $= P(\text{not TT})$ $= 1  P(\text{TT})$ $= 1  0.25 = 0.75$
$P(\text{exactly one head})$
$= P(\text{HT or TH})$ $= P(\text{HT}) + P(\text{TH})$ $= 0.25 + 0.25 = 0.50$ 
TWO FLIPS OF A FAIR COIN 
BALLS IN AN URN, WITH REPLACEMENT
Suppose that an urn contains 2 black balls, 1 red ball, and 4 green balls. Note that the sample space for this experiment is:
$S = \{\text{BB},\text{BR},\text{BG},\text{RB},\text{RR},\text{RG},\text{GB},\text{GR},\text{GG}\}$
The probability tree diagram is shown at right.
$\displaystyle
\cssId{s75}{P(\text{BB})}
\cssId{s76}{= \frac27\cdot\frac27}
\cssId{s77}{= \frac4{49}}$
$\displaystyle
\cssId{s78}{P(\text{RR})}
\cssId{s79}{= \frac17\cdot\frac17}
\cssId{s80}{= \frac1{49}}$
$\displaystyle
\cssId{s81}{P(\text{GG})}
\cssId{s82}{= \frac47\cdot\frac47}
\cssId{s83}{= \frac{16}{49}}$
$P(\text{BB or RR})$
$= P(\text{BB}) + P(\text{RR})$ $= \frac4{49} + \frac1{49}$ $= \frac5{49}$
$P(\text{at least one black ball is drawn})$
$= P(\text{BB or BR or BG or RB or GB})$ $= \frac4{49} + \frac2{49} + \frac8{49} + \frac2{49} + \frac8{49}$ $= \frac{24}{49}$
$P(\text{at most one black ball is drawn})$
$= P(\text{exactly one black ball or no black balls})$ $= P(\text{all outcomes except BB})$ $= 1  \frac4{49}$ $= \frac{45}{49}$
$P(\text{exactly one black ball is drawn})$
$= P(\text{BR or BG or RB or GB})$ $= \frac2{49} + \frac8{49} + \frac2{49} + \frac8{49}$ $= \frac{20}{49}$
$P(\text{two different color balls})$
$= P(\text{all outcomes except BB, RR, GG})$ $= 1  \frac4{49}  \frac1{49}  \frac{16}{49}$ $= \frac{28}{49}$ 
BALLS IN AN URN, WITH REPLACEMENT 
BALLS IN AN URN, WITHOUT REPLACEMENT
Suppose that an urn contains 2 black balls, 1 red ball, and 4 green balls. Note that the sample space for this experiment is:
$S = \{\text{BB},\text{BR},\text{BG},\text{RB},\text{RG},\text{GB},\text{GR},\text{GG}\}$
There is no ‘RR’ in the sample space, since if a red ball is chosen on the first draw, then there are no red balls remaining in the urn.
The probability tree diagram is shown at right.
$\displaystyle
\cssId{s119}{P(\text{BB})}
\cssId{s120}{= \frac27 \cdot \frac16}
\cssId{s121}{= \frac2{42}}
\cssId{s122}{= \frac{1}{21}}$
$P(\text{RR}) = 0$
$\displaystyle
\cssId{s124}{P(\text{GG})}
\cssId{s125}{= \frac47 \cdot \frac36}
\cssId{s126}{= \frac{12}{42}}
\cssId{s127}{= \frac{2}{7}}$
$P(\text{the same color is drawn twice})$
$= P(\text{BB or GG})$ $= \frac1{21} + \frac2{7} = \frac1{21} + \frac6{21}$ $= \frac{7}{21} = \frac13$
$\displaystyle
\cssId{s132}{P(\text{BR})}
\cssId{s133}{= \frac27 \cdot \frac16}
\cssId{s134}{= \frac2{42}}
\cssId{s135}{= \frac{1}{21}}$
$\displaystyle
\cssId{s136}{P(\text{RB})}
\cssId{s137}{= \frac17 \cdot \frac26}
\cssId{s138}{= \frac2{42}}
\cssId{s139}{= \frac{1}{21}}$
$P(\text{a black ball and a red ball are drawn})$
$= P(\text{BR or RB})$ $= \frac1{21} + \frac1{21}$ $= \frac{2}{21}$ 
BALLS IN AN URN, WITHOUT REPLACEMENT 
On this exercise, you will not key in your answer. However, you can check to see if your answer is correct. If a problem gives probabilities in decimal form, then give your answers in decimal form. If a problem gives probabilities in fraction form, then give your answers as fractions in simplest form. 
PROBLEM TYPES:
