Grade 10 Probability Venn Diagram Question - Maths...
Transcript of Grade 10 Probability Venn Diagram Question - Maths...
Grade 10 Probability Venn Diagram Question
Andy Soper
October 15, 2013
This document was constructed and type-set using PCTEX(a dielect of LATEX)
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1 Grade 10 Probability
1.1 Venn Diagram / Boolean Algebra Question
Figure 1: Venn Diagram and Boolean Algebra
(a) Complementary eventa are mutually exclusive (if one happens the othercannot happen) and exhaustive (the events cover the entire Sample Space (oneor the other must occur. There is no other possibility.)
The two events are compliments if
1. the sum of their probabilities equals 1
2. they are disjoint (no overlap)
3. the combined probabilities of the rtwo events equals 1
4. they are independent events and their probabilities are equal.
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Hence ...
1. P (A) + P (B) = 1
2. A ∩B = The empty set. Therefore A’ = B, B’ = A
3. nA+nBnS = 1
4. P (A) = P (B)
——————————————————————
(a) nA + nB = nS = 80; nA = nB = 40; x+15 = 40; x=25
(b) P(A) = 25% = 0,25; x+1580 = 0, 25; x+ 15 = 20; x=5 =⇒
(c) P (A′ ∩B′) = P (A′)× P (B′) = 15
80−x−1580 × 80−60
80 = 15 ; (65− x)× 20 = 6400
5 = 1280
1300− 20x = 1280; −20x = −20x = 1 =⇒ x=1 satisfies row 2 of (c)
(d) P (B′ ∩A) = P (A∪B)′ ⇒ P (B′)×P (A) = 1− [P (A)+P (B)−P (A)×P (B)]
80−6080 × x+15
80 = 1− [x+1580 + 60
80 −x+1580 × 60
80 ]
2080 ×
x+1580 = 1− [x+75
80 − x+1580 × 60
80 ]
2080 ×
x+1580 + [x+75
80 − x+1580 × 60
80 ] = 1
Multiply all terms by 80× 80
20(x + 15) + 80(x + 75)− 60(x + 15) = 6400
20x + 300 + 80x + 6000− 60x− 900 = 6400
40x + 5400 = 6400
40x = 1000 x = 25 =⇒ x=25 satisfies the third row of (d)
Lesson Learned: P (A′) = nS−nAnS
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Figure 2: Question2
Figure 3: Question 2 Venn Diagram
(a) n(A’) = 60 - 20 = 40 (b) n(B’) = 60 - 30 = 30 (c) n(A and B’) = 10
(d) n(A’ and B) = 20 (e) n(A or B) = 40 as they define union
(f) n(A or B’) = 20 + 20 = 40
In Progress ...
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