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Probability (4)
Conditional Probability
For these 6 balls, if 2 are chosen randomly ….
What is the probability they are a green and a red?
First ball
(from 6)
P(G) = 2/6 then ->
Second ball
(from 5)
P(R) = 1/5 P(G n R) = 2/6 x 1/5 = 1/15
P(R) = 1/6 then -> P(G) = 2/5 P(R n G) = 1/6 x 2/5 = 1/15
Either way there is a 1/15 chance of Green and Red
The 2 events are not independent
For these 6 balls, if 2 are chosen randomly ….
If the first one is Green,
what is the probability of the second being Red?
Second ball (from 5) P(R) = 1/5
This probability is CONDITIONAL on the first ball being green.
CONDITIONAL PROBABILITIES are written:- P(R|G) = 1/5
Meaning “the probability of event R given
event G has already happened”
For these 6 balls, if 2 are chosen randomly ….
If the first one is Blue,
what is the probability of the second being Red?
P(R|B) = 1/5
If the first one is Red,
what is the probability of the second being Red?
P(R|R) = 0
For these 6 balls, if 2 are chosen randomly ….
What is the probability the 2nd is Red if the 1st is Green?
First ball
(from 6)
P(G) = 2/6
Second ball
(from 5)
P(R|G) = 1/5
P(G n R) = P(G) x P(R|G) = 2/6 x 1/5 = 1/15
Since ...
P(G
Then ...
R) = P(G) x P(R|G) = P(R
P(R|G)
= P(R
G)
P(G)
G)
Conditional Probability Rule
P(B|A) = P(A B)
P(A)
“The probability that event B occurs,
given event A has occurred is :the probability they both occur
divided by the probability event A occurs”
Problem:
A math teacher gave her class two tests.
25% of the class passed both tests and 42% of the class passed the first test.
What percent of those who passed the first test also passed the second test?
Define:
F the event a pupil passed the first test
S the event a pupil passes the second test
P(F n S) = 25% = 0.25
P(F) = 42% = 0.42
P(F|S) = P(F S)
P(F)
P(F|S) = 0.25 / 0.42 = 0.60 = 60%
P(B|A) = P(A B)
P(A)
Example 1:
A jar contains black and white marbles. Two marbles are chosen without replacement.
The probability of selecting a black marble and a white marble is 0.34,
and the probability of selecting a black marble on the first draw is 0.47.
What is the probability of selecting a white marble on the second draw,
given that the first marble drawn was black?
Solution: P(White|Black)
=
P(Black and White) =
P(Black)
0.34
0.47
= 0.72
= 72%
P(B|A) = P(A B)
P(A)
Example 2:
The probability that it is Friday and that a student is absent is 0.03.
Since there are 5 school days in a week, the probability that it is Friday is 0.2.
What is the probability that a student is absent given that today is Friday?
Solution:
P(Absent|Friday)
=
P(Friday and Absent) = 0.03 = 0.15 = 15%
P(Friday)
0.2
P(B|A) = P(A B)
P(A)
Example 3:
At Kennedy Middle School, the probability that a student takes Technology and Spanish is 0.087.
The probability that a student takes Technology is 0.68.
What is the probability that a student takes Spanish given that the student is taking Technology?
Solution:
P(Spanish|Technology) =
P(Technology and Spanish)
P(Technology)
0.68
= 0.087 = 0.13
= 13%
Activity
Page 43 of your Statistics 1
book and try …
• page 43
• Exercise
B
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