A tree diagram represents the outcomes from a multi-step experiment.
A sample tree diagram is shown below.
A multi-step experiment has more than one step.
For example, here are some two-step 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 two-step experiment ‘two flips of a fair coin’. Note that the sample space for this experiment is:
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$\,S = \{\text{HH},\text{HT},\text{TH},\text{TT}\,\}\,$
The probability tree diagram is shown at right.
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$P(\text{both heads}) = P(\text{HH}) = (0.5)(0.5) = 0.25$
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$P(\text{both tails}) = P(\text{TT}) = (0.5)(0.5) = 0.25$
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$P(\text{no heads}) = P(\text{TT}) = 0.25$
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$P(\text{at least one head})$
$= P(\text{HH or HT or TH})$ $= P(\text{HH}) + P(\text{HT}) + P(\text{TH})$ $= 0.25 + 0.25 + 0.25 = 0.75$
(alternate method)
[beautiful math coming... please be patient] $P(\text{at least one head})$ $= P(\text{not TT})$ $= 1 - P(\text{TT})$ $= 1 - 0.25 = 0.75$
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$P(\text{exactly one head})$
$= P(\text{HT or TH})$ $= P(\text{HT}) + P(\text{TH})$ $= 0.25 + 0.25 = 0.50$ |
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:
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$S = \{\text{BB},\text{BR},\text{BG},\text{RB},\text{RR},\text{RG},\text{GB},\text{GR},\text{GG}\}$
The probability tree diagram is shown at right.
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$\displaystyle
P(\text{BB}) = \frac27\cdot\frac27 = \frac4{49}$
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$\displaystyle
P(\text{RR}) = \frac17\cdot\frac17 = \frac1{49}$
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$\displaystyle
P(\text{GG}) = \frac47\cdot\frac47 = \frac{16}{49}$
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$P(\text{BB or RR})$
$= P(\text{BB}) + P(\text{RR})$ $= \frac4{49} + \frac1{49}$ $= \frac5{49}$
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$P(\text{at least one black ball is drawn})$
$= P(\text{BB or BR or BG or RB or GB})$ $= \frac4{49} + \frac2{49} + \frac8{49} + \frac2{49} + \frac8{49}$ $= \frac{24}{49}$
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$P(\text{at most one black ball is drawn})$
$= P(\text{exactly one black ball or no black balls})$ $= P(\text{all outcomes except BB})$ $= 1 - \frac4{49}$ $= \frac{45}{49}$
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$P(\text{exactly one black ball is drawn})$
$= P(\text{BR or BG or RB or GB})$ $= \frac2{49} + \frac8{49} + \frac2{49} + \frac8{49}$ $= \frac{20}{49}$
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$P(\text{two different color balls})$
$= P(\text{all outcomes except BB, RR, GG})$ $= 1 - \frac4{49} - \frac1{49} - \frac{16}{49}$ $= \frac{28}{49}$ |
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:
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$S = \{\text{BB},\text{BR},\text{BG},\text{RB},\text{RG},\text{GB},\text{GR},\text{GG}\}$
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.
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$\displaystyle P(\text{BB}) = \frac27 \cdot \frac16 = \frac2{42} = \frac{1}{21}$
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$P(\text{RR}) = 0$
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$\displaystyle P(\text{GG}) = \frac47 \cdot \frac36 = \frac{12}{42} = \frac{2}{7}$
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$P(\text{the same color is drawn twice})$
$= P(\text{BB or GG})$ $= \frac1{21} + \frac2{7} = \frac1{21} + \frac6{21}$ $= \frac{7}{21} = \frac13$
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$\displaystyle P(\text{BR}) = \frac27 \cdot \frac16 = \frac2{42} = \frac{1}{21}$
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$\displaystyle P(\text{RB}) = \frac17 \cdot \frac26 = \frac2{42} = \frac{1}{21}$
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$P(\text{a black ball and a red ball are drawn})$
$= P(\text{BR or RB})$ $= \frac1{21} + \frac1{21}$ $= \frac{2}{21}$ |
BALLS IN AN URN, WITHOUT REPLACEMENT |