Conditional Probability for Intervals, Study notes of Probability and Statistics

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Intro to Contemporary Math

Conditional Probability for Intervals

Department of Mathematics UK

Announcements

I (^) A homework assignment is due next Monday. I (^) Exam 2 is next Wednesday.

Continuous Probability Reminders

Use continuous probability when picking random real numbers.

I (^) Sample spaces and events are made up of intervals. I (^) The length of an interval is the right endpoint minus the left endpoint. I (^) The probability of an interval event E is the length of E divided by the length of the sample space. I (^) The intersection of two intervals is the interval formed by their overlap.

Continuous Probability Review

Consider the sample space Ω = [ 10 , 17 ] and event (interval) F = [ 13 , 16 ]:

I (^) The sample space has length 7 − 0 = 7. I (^) Event F has length 6 − 3 = 3. I (^) Hence the probability of F is Length of F Total length

Notice that F takes up 3/7ths of the total length of the sample space.

Continuous Probability Review

Consider the sample space Ω = [ 10 , 17 ] and event (interval) F = [ 13 , 16 ]:

I (^) The sample space has length 17 − 10 = 7. I (^) Event F has length 16 − 13 = 3. I (^) Hence the probability of F is Length of F Total length

Notice that F takes up 3/7ths of the total length of the sample space.

Continuous Probability Review

Consider the sample space Ω = [ 10 , 17 ] and event (interval) F = [ 13 , 16 ]:

I (^) The sample space has length 17 − 10 = 7. I (^) Event F has length 16 − 13 = 3. I (^) Hence the probability of F is Length of F Total length

Notice that F takes up 3/7ths of the total length of the sample space.

Continuous Probability Review

Consider the sample space Ω = [ 10 , 17 ] and event (interval) F = [ 13 , 16 ]:

I (^) The sample space has length 17 − 10 = 7. I (^) Event F has length 16 − 13 = 3. I (^) Hence the probability of F is Length of F Total length

Notice that F takes up 3/7ths of the total length of the sample space.

Conditional Probability for Intervals

Let E and F be events in a sample space Ω. Then the probability of event F given that E occurred is

P(F |E ) =

Length of E

⋂ F Length of E

Conditional Probability for Intervals 1

Let Ω = [ 10 , 17 ], E = [ 11 , 16 ], and F = [ 12 , 15 ]. Let us compute P(F |E ).

I (^) Find E

⋂ F and its length:

Conditional Probability for Intervals 1

Let Ω = [ 10 , 17 ], E = [ 11 , 16 ], and F = [ 12 , 15 ]. Let us compute P(F |E ).

I (^) Find E

⋂ F and its length:

Conditional Probability for Intervals 1

Let Ω = [ 10 , 17 ], E = [ 11 , 16 ], and F = [ 12 , 15 ]. Let us compute P(F |E ).

I (^) Find E ⋂^ F and its length: E and F overlap on [ 12 , 15 ], which has length 15 − 12 = 3. I (^) Length of E is 16 − 11 = 5. I (^) Hence P(F |E ) =

Length of E

⋂ F Length of E

Notice that E

⋂ F takes up 3/5ths of the total length of E.

Conditional Probability for Intervals 1

Let Ω = [ 10 , 17 ], E = [ 11 , 16 ], and F = [ 12 , 15 ]. Let us compute P(F |E ).

I (^) Find E ⋂^ F and its length: E and F overlap on [ 12 , 15 ], which has length 15 − 12 = 3. I (^) Length of E is 16 − 11 = 5. I (^) Hence P(F |E ) =

Length of E

⋂ F Length of E

Notice that E

⋂ F takes up 3/5ths of the total length of E.

Conditional Probability for Intervals 1

Let Ω = [ 10 , 17 ], E = [ 11 , 16 ], and F = [ 12 , 15 ]. Let us compute P(F |E ).

I (^) Find E ⋂^ F and its length: E and F overlap on [ 12 , 15 ], which has length 15 − 12 = 3. I (^) Length of E is 16 − 11 = 5. I (^) Hence P(F |E ) =

Length of E

⋂ F Length of E

Notice that E

⋂ F takes up 3/5ths of the total length of E.

?(9.1) Conditional Probability Practice 1

Let Ω = [ 24 , 47 ], E = [ 29 , 43 ], and F = [ 34 , 38 ]. Compute P(F |E ). Hints:

1. Identify the intersection of [ 29 , 43 ] and [ 34 , 38 ] as an interval.
2. What is the length of the intersection?
3. What is the length of the given event?
4. Answer the question by dividing the appropriate lengths.