Probability 1 - Math Academy - JC H2 maths A levels

Probability Lesson 1

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Mr Ian Ang

c⃝ 2015 Math Academy www.MathAcademy.sg 1

Conditional Probability

If A and B are two events in our probability space, then P(A|B) is defined asthe conditional probability of A, given that B has already occured.

.Formula..

.P(A|B) =

P(A ∩ B)

P(B)




.Formula..

.P(A|B) =

P(A ∩ B)

P(B)




.Formula..

.P(A|B) =

P(A ∩ B)

P(B)


.Example (1)..

.

Suppose there are 6 red and 4 blue marbles in a bag. Two balls are drawnwithout replacement.

R

B

R

B

R

B

610

410

59

49

69

39

Q. What is P(2nd ball red)?A.

P(2nd ball red) =6

10× 5

9+

4

10× 6

9=

3

5

Q. What is P(2nd ball red | 1st ball blue)?A. We can just read the answer off from the tree diagram. Ans is 6

9.


.Example (1)..

.


R

B

R

B

R

B

610

410

59

49

69

39


P(2nd ball red) =6

10× 5

9+

4

10× 6

9=

3

5


9.


.Example (1)..

.


R

B

R

B

R

B

610

410

59

49

69

39


P(2nd ball red) =6

10× 5

9+

4

10× 6

9=

3

5


9.


.Example (1)..

.


R

B

R

B

R

B

610

410

59

49

69

39


P(2nd ball red) =6

10× 5

9+

4

10× 6

9=

3

5


9.


.Example (1)..

.


R

B

R

B

R

B

610

410

59

49

69

39


P(2nd ball red) =6

10× 5

9+

4

10× 6

9=

3

5


9.


.Example (1)..

.


R

B

R

B

R

B

610

410

59

49

69

39

Q. What is P(1st ball is blue | 2nd ball red)?A.

P(1st ball blue | 2nd ball red) =P(1st ball blue AND 2nd ball red)

P(2nd ball red)

=410

× 69

35

=4

9


.Example (1)..

.


R

B

R

B

R

B

610

410

59

49

69

39



P(2nd ball red)

=410

× 69

35

=4

9


.Example (1)..

.


R

B

R

B

R

B

610

410

59

49

69

39



P(2nd ball red)

=410

× 69

35

=4

9


.Example (2)..

.

A box contains twenty chocolates which are identical apart from their flavours.Five of the chocolates are caramel-flavoured, seven are mint-flavoured andeight are strawberry-flavoured. John randomly selects three chocolates fromthe box. Find the probability that

(a) exactly two of the three chocolates are mint-flavoured,

(b) at least two of the three chocolates have the same flavour,

(c) exactly one chocolate is strawberry-flavoured given that at least two of thethree chocolates have the same flavour.

(a) 720

× 619

× 1318

× 3!2!= 91

380

(b) 1 − P(all are different flavour)= 1 − 5

20× 7

19× 8

18× 3!

= 4357


.Example (2)..

.





(a) 720

× 619

× 1318

× 3!2!= 91

380


20× 7

19× 8

18× 3!

= 4357


.Example (2)..

.





(a) 720

× 619

× 1318

× 3!2!= 91

380


20× 7

19× 8

18× 3!

= 4357


.Example (2)..

.





(a) 720

× 619

× 1318

× 3!2!= 91

380


20× 7

19× 8

18× 3!

= 4357


.Example (2)..

.





(a) 720

× 619

× 1318

× 3!2!= 91

380


20× 7

19× 8

18× 3!

= 4357


.Example (2)..

.





(a) 720

× 619

× 1318

× 3!2!= 91

380


20× 7

19× 8

18× 3!

= 4357


.Example (2)..

.



Solution:

(c) P(exactly 1 strawberry | at least 2 same flavour)

= P(exactly 1 strawberry AND at least 2 same flavour)P(at least 2 same flavour)

P(exactly 1 strawberry AND at least 2 same flavour)= P(1 strawberry, 2 caramel) + P(1 strawberry, 2 mint)= 8

20× 5

19× 4

18× 3!

2!+ 8

20× 7

19× 6

18× 3!

2!= 62

285

∴ P(exactly 1 strawberry ∩ at least 2 same flavour)

P(at least 2 same flavour)=

622854357

=62

215


.Example (2)..

.



Solution:




20× 5

19× 4

18× 3!

2!+ 8

20× 7

19× 6

18× 3!

2!= 62

285



622854357

=62

215


.Example (2)..

.



Solution:




20× 5

19× 4

18× 3!

2!+ 8

20× 7

19× 6

18× 3!

2!= 62

285



622854357

=62

215


.Example (2)..

.



Solution:




20× 5

19× 4

18× 3!

2!+ 8

20× 7

19× 6

18× 3!

2!= 62

285



622854357

=62

215


.Example (2)..

.



Solution:




20× 5

19× 4

18× 3!

2!+ 8

20× 7

19× 6

18× 3!

2!= 62

285



622854357

=62

215


.Example (2)..

.



Solution:




20× 5

19× 4

18× 3!

2!+ 8

20× 7

19× 6

18× 3!

2!= 62

285



622854357

=62

215


.Example (2)..

.



Solution:




20× 5

19× 4

18× 3!

2!+ 8

20× 7

19× 6

18× 3!

2!= 62

285



622854357

=62

215


.Conditional Probability..

.

TYPE I“If event A happens, there is a probability of XXX that event B happens.”The above, when translated to mathematical notation, is written as

P(B|A) = XXX.

TYPE IIEG 1. Probability that a randomly chosen pear is rotten is 0.2.

P(rotten | a pear is selected ) = 0.2

EG 2. Probability that a randomly chosen male student is a prefect is 0.4.

P(prefect | a male student is selected) = 0.4



.


P(B|A) = XXX.







.


P(B|A) = XXX.







.


P(B|A) = XXX.






.Example (3)..

.

A particular diease infects a proportion p of the population of a town. A medical testhas been administered to the entire population, but unfortunately the test is notcompletely reliable. If an individual has the disease, there is a probability 0.98 ofgetting a positive result, and if an individual does not have the disease, there is aprobability of 0.08 of getting a positive result. An individual is chosen at random andtested.

Using a probability tree, find in terms of p, the probability that

(a) the individual has the disease given that the result is positive.

(b) the test will lead to a wrong conclusion.

(a) Let D be the event an individual has the disease.Let A be the event of getting a positive result.

P(D|A) = P(D ∩ A)

P(A)

=0.98p

0.98p + (1− p)0.08

=0.98p

0.9p + 0.08


.Example (3)..

.






P(D|A) = P(D ∩ A)

P(A)

=0.98p

0.98p + (1− p)0.08

=0.98p

0.9p + 0.08


.Example (3)..

.






P(D|A) = P(D ∩ A)

P(A)

=0.98p

0.98p + (1− p)0.08

=0.98p

0.9p + 0.08


.Example (3)..

.





(b)P(wrong conclusion)= P(D,A′) + P(D ′,A)= 0.02p + (1− p)0.08= 0.08 - 0.06p


.Example (3)..

.





(b)P(wrong conclusion)= P(D,A′) + P(D ′,A)= 0.02p + (1− p)0.08= 0.08 - 0.06p


.Example (4)..

.

During the monsoon season, in a particular country, each day has a probabilityof 0.5 of seeing rain. The probability that there is no traffic jam at a specificroad junction X in general is 0.32. However, on a rainy day, the probability of atraffic jam there is 0.8.Find the probability that on a randomly chosen day,

(i) there is no jam at the road junction X, given that it does not rain,

(ii) there is a traffic jam at the road junction X or it does not rain, or both,

(iii) it rains, given that there is no traffic jam at the road junction X.

[(i) 0.44 (ii) 0.9 (iii) 0.3125]


(i)

0.5

0.5

rain

no rain

jam

no jam

jam

no jam

0.8

0.2

p

1− p

P(no jam) = P(rain ∩ no jam) + P(no rain ∩ no jam)

0.32 = 0.5(0.2) + 0.5(1− p)

p = 0.56∴ P(no jam | no rain) = 1− p = 1− 0.56 = 0.44(ii) 0.5× 0.8+0.5= 0.9(iii)

P(rain | no jam) =P(rain ∩ no jam)

P(no jam)

=0.5× 0.2

0.5× 0.2 + 0.5× 0.44

=5

16c⃝ 2015 Math Academy www.MathAcademy.sg 36

.Example (5)..

.

In a local soccer tournament, the probability that Champion Football Club wins any match is 23

and the probability that it loses any match is 15 . The club is awarded three points for a win, one

point for a draw and no points for a defeat. The outcome of each match is assumed to be

independent of the other matches.

In three consecutive matches that the club plays, calculate the probability thatthe club

(a) wins more than one match, [2]

(b) obtains exactly 3 points, [3]

(c) wins only the third match, given that the club obtains exactly 3 points. [3]

[(a) 0.741 (3 s.f.) (b) 2783375

(c) 45139

]

(a) P(win 2)+P(win 3)= ( 23)2( 1

3)( 3!

2!) + ( 2

3)3 = 20

27

(b) P(win 1 and lose 2)+P(draw 3)=( 23)( 1

5)2( 3!

2!) + ( 2

15)3 = 278

3375

(c) P(win 3rd only | club contains exactly 3 pts)

=P(win 3rd only ∩ 3 pts)P(exactly 3 pts)

=P(win 3rd and lose first 2)P(exactly 3 pts)

=15× 1

5× 2

32783375

= 45139


.Example (5)..

.

In a local soccer tournament, the probability that Champion Football Club wins any match is 23

and the probability that it loses any match is 15 . The club is awarded three points for a win, one

point for a draw and no points for a defeat. The outcome of each match is assumed to be

independent of the other matches.

In three consecutive matches that the club plays, calculate the probability thatthe club

(a) wins more than one match, [2]

(b) obtains exactly 3 points, [3]

(c) wins only the third match, given that the club obtains exactly 3 points. [3]

[(a) 0.741 (3 s.f.) (b) 2783375

(c) 45139

]

(a) P(win 2)+P(win 3)= ( 23)2( 1

3)( 3!

2!) + ( 2

3)3 = 20

27

(b) P(win 1 and lose 2)+P(draw 3)=( 23)( 1

5)2( 3!

2!) + ( 2

15)3 = 278

3375

(c) P(win 3rd only | club contains exactly 3 pts)

=P(win 3rd only ∩ 3 pts)P(exactly 3 pts)

=P(win 3rd and lose first 2)P(exactly 3 pts)

=15× 1

5× 2

32783375

= 45139


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Probability 1 - Math Academy - JC H2 maths A levels