Probability 1 - Math Academy - JC H2 maths A levels
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The document discusses conditional probability and provides examples. It defines conditional probability P(A|B) as the probability of event A occurring given that event B has already occurred. An example calculates probabilities for drawing marbles from a bag. Another example finds probabilities for selecting chocolates with different flavors from a box containing chocolates of various flavors. Formulas and step-by-step workings are provided for calculating conditional probabilities.
![.
Example (4)
..
.
During the monsoon season, in a particular country, each day has a probability
of 0.5 of seeing rain. The probability that there is no traffic jam at a specific
road junction X in general is 0.32. However, on a rainy day, the probability of a
traffic 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]
c⃝ 2015 Math Academy www.MathAcademy.sg 35](https://image.slidesharecdn.com/probslides1-150422222408-conversion-gate02/85/Probability-1-Math-Academy-JC-H2-maths-A-levels-35-320.jpg)
![.
Example (5)
..
.
In a local soccer tournament, the probability that Champion Football Club wins any match is 2
3
and the probability that it loses any match is 1
5 . 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 that
the 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) 278
3375
(c) 45
139
]
(a) P(win 2)+P(win 3)= (2
3
)2
(1
3
)(3!
2!
) + (2
3
)3
= 20
27
(b) P(win 1 and lose 2)+P(draw 3)=(2
3
)(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)
=
1
5
× 1
5
× 2
3
278
3375
= 45
139
c⃝ 2015 Math Academy www.MathAcademy.sg 37](https://image.slidesharecdn.com/probslides1-150422222408-conversion-gate02/85/Probability-1-Math-Academy-JC-H2-maths-A-levels-37-320.jpg)
![.
Example (5)
..
.
In a local soccer tournament, the probability that Champion Football Club wins any match is 2
3
and the probability that it loses any match is 1
5 . 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 that
the 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) 278
3375
(c) 45
139
]
(a) P(win 2)+P(win 3)= (2
3
)2
(1
3
)(3!
2!
) + (2
3
)3
= 20
27
(b) P(win 1 and lose 2)+P(draw 3)=(2
3
)(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)
=
1
5
× 1
5
× 2
3
278
3375
= 45
139
c⃝ 2015 Math Academy www.MathAcademy.sg 38](https://image.slidesharecdn.com/probslides1-150422222408-conversion-gate02/85/Probability-1-Math-Academy-JC-H2-maths-A-levels-38-320.jpg)
Probability 1 - Math Academy - JC H2 maths A levels
- 1. Probability Lesson 1 www.MathAcademy.sg Mr Ian Ang c⃝ 2015 Math Academy www.MathAcademy.sg 1
- 2. Conditional Probability If A and B are two events in our probability space, then P(A|B) is defined as the conditional probability of A, given that B has already occured. . Formula .. . P(A|B) = P(A ∩ B) P(B) c⃝ 2015 Math Academy www.MathAcademy.sg 2
- 3. Conditional Probability If A and B are two events in our probability space, then P(A|B) is defined as the conditional probability of A, given that B has already occured. . Formula .. . P(A|B) = P(A ∩ B) P(B) c⃝ 2015 Math Academy www.MathAcademy.sg 3
- 4. Conditional Probability If A and B are two events in our probability space, then P(A|B) is defined as the conditional probability of A, given that B has already occured. . Formula .. . P(A|B) = P(A ∩ B) P(B) c⃝ 2015 Math Academy www.MathAcademy.sg 4
- 5. . Example (1) .. . Suppose there are 6 red and 4 blue marbles in a bag. Two balls are drawn without replacement. R B R B R B 6 10 4 10 5 9 4 9 6 9 3 9 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 . c⃝ 2015 Math Academy www.MathAcademy.sg 5
- 6. . Example (1) .. . Suppose there are 6 red and 4 blue marbles in a bag. Two balls are drawn without replacement. R B R B R B 6 10 4 10 5 9 4 9 6 9 3 9 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 . c⃝ 2015 Math Academy www.MathAcademy.sg 6
- 7. . Example (1) .. . Suppose there are 6 red and 4 blue marbles in a bag. Two balls are drawn without replacement. R B R B R B 6 10 4 10 5 9 4 9 6 9 3 9 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 . c⃝ 2015 Math Academy www.MathAcademy.sg 7
- 8. . Example (1) .. . Suppose there are 6 red and 4 blue marbles in a bag. Two balls are drawn without replacement. R B R B R B 6 10 4 10 5 9 4 9 6 9 3 9 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 . c⃝ 2015 Math Academy www.MathAcademy.sg 8
- 9. . Example (1) .. . Suppose there are 6 red and 4 blue marbles in a bag. Two balls are drawn without replacement. R B R B R B 6 10 4 10 5 9 4 9 6 9 3 9 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 . c⃝ 2015 Math Academy www.MathAcademy.sg 9
- 10. . Example (1) .. . Suppose there are 6 red and 4 blue marbles in a bag. Two balls are drawn without replacement. R B R B R B 6 10 4 10 5 9 4 9 6 9 3 9 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) = 4 10 × 6 9 3 5 = 4 9 c⃝ 2015 Math Academy www.MathAcademy.sg 10
- 11. . Example (1) .. . Suppose there are 6 red and 4 blue marbles in a bag. Two balls are drawn without replacement. R B R B R B 6 10 4 10 5 9 4 9 6 9 3 9 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) = 4 10 × 6 9 3 5 = 4 9 c⃝ 2015 Math Academy www.MathAcademy.sg 11
- 12. . Example (1) .. . Suppose there are 6 red and 4 blue marbles in a bag. Two balls are drawn without replacement. R B R B R B 6 10 4 10 5 9 4 9 6 9 3 9 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) = 4 10 × 6 9 3 5 = 4 9 c⃝ 2015 Math Academy www.MathAcademy.sg 12
- 13. . 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 and eight are strawberry-flavoured. John randomly selects three chocolates from the 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 the three chocolates have the same flavour. (a) 7 20 × 6 19 × 13 18 × 3! 2! = 91 380 (b) 1 − P(all are different flavour) = 1 − 5 20 × 7 19 × 8 18 × 3! = 43 57 c⃝ 2015 Math Academy www.MathAcademy.sg 13
- 14. . 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 and eight are strawberry-flavoured. John randomly selects three chocolates from the 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 the three chocolates have the same flavour. (a) 7 20 × 6 19 × 13 18 × 3! 2! = 91 380 (b) 1 − P(all are different flavour) = 1 − 5 20 × 7 19 × 8 18 × 3! = 43 57 c⃝ 2015 Math Academy www.MathAcademy.sg 14
- 15. . 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 and eight are strawberry-flavoured. John randomly selects three chocolates from the 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 the three chocolates have the same flavour. (a) 7 20 × 6 19 × 13 18 × 3! 2! = 91 380 (b) 1 − P(all are different flavour) = 1 − 5 20 × 7 19 × 8 18 × 3! = 43 57 c⃝ 2015 Math Academy www.MathAcademy.sg 15
- 16. . 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 and eight are strawberry-flavoured. John randomly selects three chocolates from the 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 the three chocolates have the same flavour. (a) 7 20 × 6 19 × 13 18 × 3! 2! = 91 380 (b) 1 − P(all are different flavour) = 1 − 5 20 × 7 19 × 8 18 × 3! = 43 57 c⃝ 2015 Math Academy www.MathAcademy.sg 16
- 17. . 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 and eight are strawberry-flavoured. John randomly selects three chocolates from the 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 the three chocolates have the same flavour. (a) 7 20 × 6 19 × 13 18 × 3! 2! = 91 380 (b) 1 − P(all are different flavour) = 1 − 5 20 × 7 19 × 8 18 × 3! = 43 57 c⃝ 2015 Math Academy www.MathAcademy.sg 17
- 18. . 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 and eight are strawberry-flavoured. John randomly selects three chocolates from the 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 the three chocolates have the same flavour. (a) 7 20 × 6 19 × 13 18 × 3! 2! = 91 380 (b) 1 − P(all are different flavour) = 1 − 5 20 × 7 19 × 8 18 × 3! = 43 57 c⃝ 2015 Math Academy www.MathAcademy.sg 18
- 19. . 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 and eight are strawberry-flavoured. John randomly selects three chocolates from the box. Find the probability that (c) exactly one chocolate is strawberry-flavoured given that at least two of the three chocolates have the same flavour. 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) = 62 285 43 57 = 62 215 c⃝ 2015 Math Academy www.MathAcademy.sg 19
- 20. . 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 and eight are strawberry-flavoured. John randomly selects three chocolates from the box. Find the probability that (c) exactly one chocolate is strawberry-flavoured given that at least two of the three chocolates have the same flavour. 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) = 62 285 43 57 = 62 215 c⃝ 2015 Math Academy www.MathAcademy.sg 20
- 21. . 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 and eight are strawberry-flavoured. John randomly selects three chocolates from the box. Find the probability that (c) exactly one chocolate is strawberry-flavoured given that at least two of the three chocolates have the same flavour. 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) = 62 285 43 57 = 62 215 c⃝ 2015 Math Academy www.MathAcademy.sg 21
- 22. . 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 and eight are strawberry-flavoured. John randomly selects three chocolates from the box. Find the probability that (c) exactly one chocolate is strawberry-flavoured given that at least two of the three chocolates have the same flavour. 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) = 62 285 43 57 = 62 215 c⃝ 2015 Math Academy www.MathAcademy.sg 22
- 23. . 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 and eight are strawberry-flavoured. John randomly selects three chocolates from the box. Find the probability that (c) exactly one chocolate is strawberry-flavoured given that at least two of the three chocolates have the same flavour. 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) = 62 285 43 57 = 62 215 c⃝ 2015 Math Academy www.MathAcademy.sg 23
- 24. . 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 and eight are strawberry-flavoured. John randomly selects three chocolates from the box. Find the probability that (c) exactly one chocolate is strawberry-flavoured given that at least two of the three chocolates have the same flavour. 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) = 62 285 43 57 = 62 215 c⃝ 2015 Math Academy www.MathAcademy.sg 24
- 25. . 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 and eight are strawberry-flavoured. John randomly selects three chocolates from the box. Find the probability that (c) exactly one chocolate is strawberry-flavoured given that at least two of the three chocolates have the same flavour. 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) = 62 285 43 57 = 62 215 c⃝ 2015 Math Academy www.MathAcademy.sg 25
- 26. . 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 II EG 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 c⃝ 2015 Math Academy www.MathAcademy.sg 26
- 27. . 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 II EG 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 c⃝ 2015 Math Academy www.MathAcademy.sg 27
- 28. . 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 II EG 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 c⃝ 2015 Math Academy www.MathAcademy.sg 28
- 29. . 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 II EG 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 c⃝ 2015 Math Academy www.MathAcademy.sg 29
- 30. . Example (3) .. . A particular diease infects a proportion p of the population of a town. A medical test has been administered to the entire population, but unfortunately the test is not completely reliable. If an individual has the disease, there is a probability 0.98 of getting a positive result, and if an individual does not have the disease, there is a probability of 0.08 of getting a positive result. An individual is chosen at random and tested. 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 c⃝ 2015 Math Academy www.MathAcademy.sg 30
- 31. . Example (3) .. . A particular diease infects a proportion p of the population of a town. A medical test has been administered to the entire population, but unfortunately the test is not completely reliable. If an individual has the disease, there is a probability 0.98 of getting a positive result, and if an individual does not have the disease, there is a probability of 0.08 of getting a positive result. An individual is chosen at random and tested. 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 c⃝ 2015 Math Academy www.MathAcademy.sg 31
- 32. . Example (3) .. . A particular diease infects a proportion p of the population of a town. A medical test has been administered to the entire population, but unfortunately the test is not completely reliable. If an individual has the disease, there is a probability 0.98 of getting a positive result, and if an individual does not have the disease, there is a probability of 0.08 of getting a positive result. An individual is chosen at random and tested. 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 c⃝ 2015 Math Academy www.MathAcademy.sg 32
- 33. . Example (3) .. . A particular diease infects a proportion p of the population of a town. A medical test has been administered to the entire population, but unfortunately the test is not completely reliable. If an individual has the disease, there is a probability 0.98 of getting a positive result, and if an individual does not have the disease, there is a probability of 0.08 of getting a positive result. An individual is chosen at random and tested. 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. (b) P(wrong conclusion) = P(D, A′ ) + P(D′ , A) = 0.02p + (1 − p)0.08 = 0.08 - 0.06p c⃝ 2015 Math Academy www.MathAcademy.sg 33
- 34. . Example (3) .. . A particular diease infects a proportion p of the population of a town. A medical test has been administered to the entire population, but unfortunately the test is not completely reliable. If an individual has the disease, there is a probability 0.98 of getting a positive result, and if an individual does not have the disease, there is a probability of 0.08 of getting a positive result. An individual is chosen at random and tested. 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. (b) P(wrong conclusion) = P(D, A′ ) + P(D′ , A) = 0.02p + (1 − p)0.08 = 0.08 - 0.06p c⃝ 2015 Math Academy www.MathAcademy.sg 34
- 35. . Example (4) .. . During the monsoon season, in a particular country, each day has a probability of 0.5 of seeing rain. The probability that there is no traffic jam at a specific road junction X in general is 0.32. However, on a rainy day, the probability of a traffic 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] c⃝ 2015 Math Academy www.MathAcademy.sg 35
- 36. (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
- 37. . Example (5) .. . In a local soccer tournament, the probability that Champion Football Club wins any match is 2 3 and the probability that it loses any match is 1 5 . 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 that the 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) 278 3375 (c) 45 139 ] (a) P(win 2)+P(win 3)= (2 3 )2 (1 3 )(3! 2! ) + (2 3 )3 = 20 27 (b) P(win 1 and lose 2)+P(draw 3)=(2 3 )(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) = 1 5 × 1 5 × 2 3 278 3375 = 45 139 c⃝ 2015 Math Academy www.MathAcademy.sg 37
- 38. . Example (5) .. . In a local soccer tournament, the probability that Champion Football Club wins any match is 2 3 and the probability that it loses any match is 1 5 . 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 that the 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) 278 3375 (c) 45 139 ] (a) P(win 2)+P(win 3)= (2 3 )2 (1 3 )(3! 2! ) + (2 3 )3 = 20 27 (b) P(win 1 and lose 2)+P(draw 3)=(2 3 )(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) = 1 5 × 1 5 × 2 3 278 3375 = 45 139 c⃝ 2015 Math Academy www.MathAcademy.sg 38