Probability Assignment #41 Review Paper #1 day7

Probability A#41 Review Paper #1 day7

5. [ Maximum mark: 6] Let A and B be independent events, where P(A) = 0.6 and P(B) = x. (a)Write down an expression for P(A  B). (b) Given that P(A  B) = 0.8 (i) Find x; (ii) Find P(A  B) (c) Hence, explain why A and B are not mutually exclusive. 2008/SP1/TZ0 Complement Combined events Mutually exclusive events Conditional Independent events

Defn of Independent events: P(A  B) = P(A)  P(B) 2008/SP1/TZ0 (a) P(A  B) = P(A)  P(B) P(A  B) = (0.6)  ( x ) = 0.6 x x 1 – x P(A) = P(B) = A B B’ A’ B B’

Defn of Independent events: P(A  B) = P(A)  P(B) 2008/SP1/TZ0 Independent: P(A  B) = P(A) + P(B) – P(A)  P(B) P(A  B) = P(A) + P(B) – P(A  B) (b) (i) 0.8 = x – (0.6)( x ) 0.2 = 0.4 x x = 0.5 (b)(ii) P(A  B) = P(A)  P(B) = (0.6)(0.5) = 0.3 (c) A and B are not mutually exclusive because P(A  B)  0.

07/5/MATME/SP1/ENG/TZ1/ P(A  B) = P(A) + P(B) – P(A  B) Complement Combined events Mutually exclusive events Conditional Independent events

Conditional probability: P(A  B) = P(A)  P(B|A) 07/5/MATME/SP1/ENG/TZ1/Conditional probability (a) (b) P(B) = 1 – P ’ (B) (c) 3SF

2008/SP1/TZ0/Section B Let X be the sum of the scores on the two dice. (a) Find Two standard six-sided dice are tossed. A diagram representing the sample space is shown below. Find the value of k for which Elena’s expected number of points is zero (b)

2008/SP1/TZ0/ Section B +Die B A (a) (i) number of ways of getting X = 6 is 5 (ii) number of ways of getting X > 6 is 21 (iii) Probability Conditional probability: P(A  B) = P(A)  P(B|A)

Probability 2008/SP1/TZ0/ Section B +Die B A (b) number of ways of getting X < 6 is 10 Probability

Expectation or mean value or expected mean 2008/SP1/TZ0/ Section BProbability x < 6x = 6x > 6 w–k–k31 P(W = w) Sum = 1 Probability distribution table

Probability Assignment #42 Review Paper #1 day8

08/5/MATME/SP1/ENG/TZ2 ans

Probability08/5/MATME/SP1/ENG/TZ2 (a)(i) P(male  T) = P(male) + P(T) – P(Male  T) P(male or Tennis) (a) (ii) P(not football | female) Conditional probability: P(A  B) = P(A)  P(B|A) ?

Fb’ Fb Fb’ Fb Conditional probability Probability08/5/MATME/SP1/ENG/TZ2 (b) 0.4 P(1 st Fb’) = P(2 nd Fb’|1 st Fb’) = ?

2010/SP1/TZ1/ Consider the events A and B, where P(A) = 0.5, P(B) = 0.7 and P(A  B) =0.3. The Venn diagram below shows the events A and B, and the probabilities p, q, and r. answers

Probability (a) (i) p = 0.2 (ii) q = 0.4 (iii) r = 0.1 (b) P(A|B ’ ) = (c) Independent  P(A  B) = P(A)  P(B) P(A  B) = 0.3 P(A)P(B) = (0.5)(0.7) =  0.35 so A and B are not independent. 2010/SP1/TZ1/.3 Defn of Independent events: P(A  B) = P(A)  P(B) = =.4 1 – ( ) =.1 ?

08/SP1/TZ1/ Section B ans

08/SP1/TZ1/ Section B ans Find the value of p, of s and of t.

Intersection Probability08/SP1/TZ1/ Section B (a) (b) (c) (i) Score = 2 Score = 3 Score = 2 ?

08/SP1/TZ1/ Section B ans

Intersection & union Probability08/SP1/TZ1/ Section B (c) (ii) Score = 2 Score = 3 Score = 2 ?

Probability08/SP1/TZ1/ Section B (d) (i) Probability distribution table x23 P(X = x) Sum = 1 (d) (ii) Expectation or mean value or expected mean ?

Probability08/SP1/TZ1/ Section B (e) List all possible outcomes Game 1 Score Game 2 Score Total $Probability 22$0 23$ $20 ?

2010/SP1/TZ2 ans

Intersection & union Probability (a) (i) P(E  F) 2010/SP1/TZ2 P(E ’  F) (a) (ii) P(F) = P(E  F) + P(E ’  F) ?

Probability 2010/SP1/TZ2 (b) (i) P(E  F ’ ) (b) (ii) P(E | F) = Conditional probability: P(A  B) = P(A)  P(B|A) ?

Probability (c) List all possible outcomes Monday’s Bus Tuesday’s Bus Total $Probability Miss (E) $0 Miss (E) Take (E ’ ) $3 Take (E ’ ) Miss (E)$3 Take (E ’ ) $6 2010/SP1/TZ2 ?

Probability (c) Probability distribution table x (cost in euros) 036 P(X ) Sum = 1 (d) Expectation or mean value or expected mean 2010/SP1/TZ2 ?

09/SP1/TZ0 Section B ans

09/SP1/TZ0 Section B

Probability Venn diagram: Union & intersection ? Neither: 0 both ? A: Football 55 B : Rugby 75 Universe: 100 boys 09/SP1/TZ0 Section B ? 100 = – n(A  B) A  BA  B (a) (i) n(A  B) = n(A) + n(B) – n(A  B) 100 = 130 – n(A  B) n(A  B) = 130 – 100 = 30 boys play both (a) (ii) 45 boys play only rugby n(B) – n(A  B) = 75 – 30 = 45 ?

Probability Venn diagram 45 Neither: 0 ABAB ? A: Football 55 B : Rugby 75 Universe: 100 boys 09/SP1/TZ0 Section B (b) (i) 25 boys play only football n(A) – n(A  B) = 55 – 30 = 25 n(Boys who play only 1 sport) = 100 – (boys who play both) = 100 – 30 = 70 P(he plays only 1 sport) = = ?

Universe: 100 boys who play only 1 sport 70 Probability 45 Neither: 0 ABAB A: Football 55 B : Rugby 75 Universe: 100 boys 09/SP1/TZ0 Section B (b) (ii) P(Rugby | he plays only 1 sport) “given” Conditional probability: P(A  B) = P(A)  P(B|A) 2545 ?

Probability 45 Neither: 0 ABAB A: Football 55 B : Rugby 75 Universe: 100 boys 09/SP1/TZ0 Section B (c) Events A and B are mutually exclusive iff P(A  B) is zero. So events A and B are not mutually exclusive P(A  B) = 0.30  0 Mutually exclusive events: P(A  B) = 0 ?

Probability 45 Neither: 0 ABAB A: Football 55 B : Rugby 75 Universe: 100 boys 09/SP1/TZ0 Section B (d) Events A and B are independent iff P(A  B) = P(A)  P(B) 0.30  (0.55)  (0.75) 0.30  So events A and B are not independent Defn of Independent events: P(A  B) = P(A)  P(B) ?

End of Probability slides

09/SP1/TZ0 Section B

Statistics

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STATISTICS M08/5/MATME/SP1/ENG/TZ2/

Statistics 07/5/MATME/SP1/ENG/TZ1

Statistics – Calculator 2007/SP1/ENG/TZ1/