when a normal, unbiased, 6-sided die is thrown, the numbers 1 to 6 are possible. these are the...
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Taken together, are these events exhaustive or not? Throw a die and get an odd number Throw a die and get an even number {1, 3, 5} {2, 4, 6}TRANSCRIPT
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S1PROBABILITY
Laws ofProbability
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Exhaustive eventsWhen a normal, unbiased, 6-sided die is thrown, the numbers 1 to 6 are possible.These are the ONLY ‘events’ possible.This means these are EXHAUSTIVE as they cover ALL POSSIBLE OUTCOMES.
The SUM of the probabilities of a set of EXHAUSTIVE EVENTS is ALWAYS 1
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Exhaustive eventsTaken together, are these events exhaustive or not?
A BThrow a die and get an odd number
Throw a die and get an even number
{1, 3, 5} {2, 4, 6}
Yes
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Exhaustive eventsTaken together, are these events exhaustive or not?
A BGet a 1 Get a number greater
than 2
{1} {3, 4, 5, 6}
No
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Exhaustive eventsTaken together, are these events exhaustive or not?
A BThrow a die and get a prime number
Throw a die and get a composite number
{2, 3, 5} {4, 6}
No
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Complimentary eventsEvent A:Probability of getting a 1 on a die = P(1) = 1/6
Event B:Probability of NOT getting a 1 on a die = P(1’) = 5/6
Event B is the COMPLIMENT of event A
An event and its compliment are EXHAUSTIVE
P(A’) = 1 – P(A)
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Mutually Exclusive eventsEvent A:Throw a die and get an EVEN NUMBER
Event B:Throw a die and get an ODD NUMBER
Events A and B CANNOT HAPPEN AT THE SAME TIME this means that events A and B are MUTUALLY EXCLUSIVE
If two events, A and B, are MUTUALLY EXCLUSIVE then…
P(A B) = 0
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Independent Events
A coin is thrown and it lands on HEADS.
It is thrown again. Is the probability of getting a head the 2nd time greater, less or the same?
The same coin is thrown 6 times and each time ends up as a head. If it is thrown for a 7th time is the probability of getting ANOTHER head the 7th time greater, less or the same this time?
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Independent EventsIf two events A and B are INDEPENDENT it means that if A happens then THIS DOES NOT AFFECT the probability of B occurring.
P(A B) = P(A) X P(B)
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Addition Rule
A B
P(A B) = P(A) + P(B) – P(A B)
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Conditional Probability
A B
P(A | B) =The probability
of A GIVEN THAT B has already
happened.
P(B)P(A B)
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Conditional Probability
P(A | B) =P(B)
P(A B)
If A and B are MUTUALLY EXCLUSIVE, P(A B) = 0
P(A | B) = 0If A and B are INDEPENDENT, P(A B) = P(A) X P(B)
P(A | B) =P(B)
P(A) X P(B)
P(A | B) = P(A)
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The events A and B are such that P(A) = 5/16 , P(B) = ½ , P(A|B) = ¼ Find(a) P(A B) (2)(b) P(B’ | A) (3)(c) P(A’ B)
(2)(d) Determine, with a reason, whether or not events A and B are independent
(3)
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The events A and B are such that P(A) = 5/16 , P(B) = ½ , P(A|B) = ¼ Find(a) P(A B) (2)
𝑷 ( 𝑨|𝑩 )=𝑷 (𝑨∩𝑩)𝑷 (𝑩)
𝑷 ( 𝑨|𝑩 )×𝑷 (𝑩)=𝑷 (𝑨∩𝑩)
𝟏𝟒×
𝟏𝟐=𝑷 (𝑨∩𝑩)
𝑷 ( 𝑨∩𝑩)=𝟏𝟖
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The events A and B are such that P(A) = 5/16 , P(B) = ½ , P(A|B) = ¼ Find(b) P(B’ | A) (3)
𝑷 (𝑩 ′|𝑨)=𝑷 (𝑩 ′ ∩𝑨)𝑷 (𝑨)
A B 𝑷 (𝑩′∩𝑨 )=𝑷 ( 𝑨)−𝑷 (𝑨∩𝑩)
𝑷 (𝑩′∩𝑨 )= 𝟓𝟏𝟔 −
𝟏𝟖¿
𝟑𝟏𝟔
𝑷 (𝑩 ′|𝑨)=
𝟑𝟏𝟔𝟓𝟏𝟔
¿𝟑𝟓
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The events A and B are such that P(A) = 5/16 , P(B) = ½ , P(A|B) = ¼ Find(c) P(A’ B) (2)
𝑷 ( 𝑨′∪𝑩)=𝑷 ( 𝑨 ′ )+𝑷 (𝑩 )−𝑷 (𝑨′∩𝑩)
𝑷 ( 𝑨′∪𝑩)=¿(𝟏− 𝟓𝟏𝟔 )+𝟏𝟐 −(𝟏𝟐−𝟏𝟖 )
𝑷 ( 𝑨′∪𝑩)=𝟏𝟏𝟏𝟔 +
𝟖𝟏𝟔−
𝟔𝟏𝟔
𝑷 ( 𝑨′∪𝑩)=𝟏𝟑𝟏𝟔
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The events A and B are such that P(A) = 5/16 , P(B) = ½ , P(A|B) = ¼ Find(d) Determine, with a reason, whether or not events A and B are independent
(3)
If A and B are independent then P(A B) = P(A) X P(B)
𝑷 ( 𝑨∩𝑩)=𝟏𝟖
𝑷 (𝑨)×𝑷 (𝑩 )= 𝟓𝟏𝟔 ×
𝟏𝟐¿
𝟓𝟑𝟐≠
𝟏𝟖
NOT independent
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The events A and B are such that P(A) = 0.2 and P(A B) = 0.6 Find(a) P(A’ B’)
(2)(b) P(A’ B)
(2)Given also that events A and B are independent, find(c) P(B)
(4)(d) P(A’ B’)
(2)
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A B
The events A and B are such that P(A) = 0.2 and P(A B) = 0.6 Find(a) P(A’ B’)
(2)
𝑷 ( 𝑨′∩𝑩 ′ )=¿𝟏−𝑷 (𝑨∪𝑩)
𝑷 ( 𝑨′∩𝑩 ′ )=¿𝟏−𝟎 .𝟔¿𝟎 .𝟒
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The events A and B are such that P(A) = 0.2 and P(A B) = 0.6 Find(b) P(A’ B) (2)
A BA B
𝑷 ( 𝑨′∩𝑩 )=¿𝑷 ( 𝑨∪𝑩)−𝑷 ( 𝑨)
𝑷 ( 𝑨′∩𝑩 )=𝟎.𝟔−𝟎 .𝟐𝑷 ( 𝑨′∩𝑩 )=𝟎.𝟒
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The events A and B are such that P(A) = 0.2 and P(A B) = 0.6 FindGiven also that events A and B are independent, find(c) P(B)
(4)𝑷 ( 𝑨∪𝑩)=𝑷 ( 𝑨 )+𝑷 (𝑩 )−𝑷 (𝑨∩𝑩)
𝑷 ( 𝑨∩𝑩)=𝑷 (𝑨)×𝑷 (𝑩)𝑷 ( 𝑨∪𝑩)=𝑷 ( 𝑨 )+𝑷 (𝑩 )−𝑷 ( 𝑨 )𝑷 (𝑩)
𝟎 .𝟔=𝟎 .𝟐+𝑷 (𝑩 )−𝟎 .𝟐𝑷 (𝑩)
𝟎 .𝟔=𝟎 .𝟐+𝟎 .𝟖𝑷 (𝑩 )𝟎 .𝟒=𝟎 .𝟖𝑷 (𝑩 )𝟎 .𝟒𝟎 .𝟖=𝑷 (𝑩 )¿𝟎 .𝟓
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The events A and B are such that P(A) = 0.2 and P(A B) = 0.6 FindGiven also that events A and B are independent, find(d) P(A’ B’) (2)
𝑷 ( 𝑨 ′∪𝑩 ′ )=𝑷 ( 𝑨 ′ )+𝑷 (𝑩 ′ )−𝑷 (𝑨 ′ ∩𝑩 ′)𝑷 ( 𝑨 ′∪𝑩 ′ )=𝟎 .𝟖+𝟎 .𝟓−𝟎 .𝟒𝑷 ( 𝑨 ′∪𝑩 ′ )=𝟎 .𝟗