# Probability with tree diagram - Excel by tbp20087

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```									                                            PROBABILITY USING TREE DIAGRAMS
6        # Consider the sample space when three coins are tossed:

H HHH
H                 T HHT
H
n(S) = 8                                                T                 H HTH
T HTT
0
H THH
H
T THT
T
T                 H TTH
T TTT
Coin 1            Coin 2        Coin 3

Find the following probabilities
P(HHT / last two coins show HH)

r
◄                                                                                 1
Random no                                                                         2
3
4
LOOKUP("C",{"a","b","c","d";1,2,3,4}) equals 3     5
6
Gives qu no.                    ####                                              7
Writes question                 ####                                              8
9
No, Try again.                                       12
P(HHT / last two coins show HH)                 13
P(HHT / last two coins show HH)
Equivalent expressions                                                           14
P(HHT / last two coins show HH)                               15
16
17
1   Correct. Well done.   4   18
2   Yes!                  5   19
3   Right!                    20
4   Well done!                21
5   Correct.                  22
23
1   No!                       24
2   Try again.                25
3   Not correct!
4   Have another try!
5   Wrong!
SING TREE DIAGRAMS
ce when three coins are tossed:
Possible outcomes

H HHH           *
H                T HHT           *

T                H HTH           *
T HTT           *

H THH           *
H
T THT           *
H TTH           *
T
T TTT           *
Coin 3

0

d
s

P(HHH)                             0.125    a
P(TTT)                             0.125    b
P(THT)                             0.125    c
P(at least one T)                  0.875    d
P(coin 1 = H)                         0.5   e
P(exactly two T)                   0.375    f
P(exactly two H)                   0.375    g
P(exactly one H)                   0.375    h
P(the last toss is a H)               0.5   I
P(TTT / at least two T)             0.25    j
P(TTT / at least one T)            0.143    k
P(TTT / at least two H)                 0   l
P(THH / at least one H)            0.143    m
P( last two tosses are T)           0.25    n
P( first two tosses are T)          0.25    o
P( last two tosses are H)          0.25   p
P( first two tosses are H)         0.25   q
P(at least one H)                 0.875   r
P(at least two H / at least one H)0.571   s
P(at least two T / at least one T)0.571   t
P(HHT / at least one H)           0.143   u
P(HHH / first two coins show HH)0.5       v
P(HHT / last two coins show HH) 0         x
P(HHH / at least one H)           0.143   y
P(HTH / at least two H)            0.25   z
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PROBABILITY USING TREE DIAGRAMS
24   # Assume that 5 % of males and 1 % of females are colourblind.
POSSIBLE
OUTCOMES                     PROBABILITIES
0.05             C      MC         P(MC) = O.5 x 0.05 = 0.025
M
0.5                  0.95              N      MN          P(MN) = O.5 x 0.95 = 0.475
0

0.5                    0.01             C     FC         P(FC) = O.5 x 0.01 = 0.005
F
0.99            N      FN        P(FN) = O.5 x 0.99 = 0.495

Tree diagram
A person is chosen at random and tested for colourblindness.

Find the following probabilities
P(colourblind / female)

1
Random no                                                                         2
3
4
LOOKUP("C",{"a","b","c","d";1,2,3,4}) equals 3    5
6
Gives qu no.                     10                                               7
Writes question                 P(colourblind / female)                           8
9
Correct. Well done.
P(colourblind / female)
Equivalent expressions P(colourblind / female)
P(colourblind / female)
1   Correct. Well done.   5
2   Yes!                  4
3   Right!
4   Well done!
5   Correct.

1   No!
2   Try again.
3   Not correct!
4   Have another try!
5   Wrong!
SING TREE DIAGRAMS
es and 1 % of females are colourblind.
E
ES         PROBABILITIES
Colourblind Normal
P(MC) = O.5 x 0.05 = 0.025
Male            0.025       0.475      0.5
P(MN) = O.5 x 0.95 = Female
0.475                       0.005       0.495      0.5
Totals           0.03        0.97       1

P(FC) = O.5 x 0.01 = 0.005
Two way table
P(FN) = O.5 x 0.99 = 0.495
It is sometimes easier to use the information
in a two way table.
d tested for colourblindness.

0.01

d
s

P(a male)                                0.5
P(a female)                              0.5
P(a colourblind female)               0.005
P(a colourblind male)                 0.025
P(a colourblind person)                0.03
P(a person who is not colourblind)     0.97
P(female / colourblind)          0.1666667
P(male / colourblind)            0.8333333
P(colourblind / male)                  0.05
P(colourblind / female)                0.01

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