When a normal, unbiased, 6-sided die is thrown, the numbers 1 to 6 are possible. These are the...

S1PROBABILITY

Laws ofProbability

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

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

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

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

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)

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

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?

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)

Addition Rule

A B

P(A B) = P(A) + P(B) – P(A B)

Conditional Probability

A B

P(A | B) =The probability

of A GIVEN THAT B has already

happened.

P(B)P(A B)

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)

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)

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

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

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

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

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)

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

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

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

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