O lvl E Maths: Probability

Question from http://sgforums.com/forums/2297/topics/345310

In a city, 20% of the citizens are carriers of a certain disease.If a citizen is a carrier, the probabbility that a blood test will indicate a positive result is 0.95. If citizen is a non-carrier, the probability that the blood test will indicate a postive result is 0.10. Two citizens are selected at random for the blood test. Find the probability that

(a) both citizens have positive results
(b) both citizens have negative results

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Answer:


(a) P(positive results) = 0.2 * 0.95 + 0.8 * 0.1 = 0.27
P(both citizens have positive results)
= P(positive results) * P(positive results)
= 0.27 * 0.27
= 0.0729

(b) P(negative results) = 0.2 * 0.05 + 0.8 * 0.9 = 0.73
P(both citizens have negative results)
= P(negative results) * P(negative results)
= 0.73 * 0.73
= 0.5329

O lvl E Maths: Probability

Question from http://sgforums.com/forums/2297/topics/344458

The probabilities that Alif, Kannan and Caili will pass an art examination are 4/5, 5/7, and 2/3 respectively. Find the probability that

(a) at least 1 of them will pass the art examination

(b) Only Alif will pass the art examination

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Answer:

a) P(all fail) = 1/5 * 2/7 * 1/3 = 2/105

P(at least one pass) = 1 - 2/105 = 103/105

b) P(only Alif will pass the art examination) = 4/5 * 2/7 * 1/3 = 8/105

O lvl E Maths: Probability

From http://www.sgforums.com/forums/2297/topics/348803

Suppose that 45% and 40% of Singaporeans approve and disapprove the idea of building a casino respectively. If three Singaporeans are interviewed at random,find the probability that
a) all of them will either approve or disapprove of the idea

b) at least two of them will not approve of the idea

c) at most one of them will approve of the idea.

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Answer:

Let A = Approve, D = disapprove

(a) P(all of them have opinions) = 0.85 * 0.85 * 0.85 = 4913/8000

(b) Let X = not disapprove
P(X) = 1 - 0.4 = 0.6

P(at least 2 disapprove) = P(DDX) + P(DXD) + P(XDD) + P(DDD)
= 3 (0.4 * 0.4 * 0.6) + 0.4 * 0.4 * 0.4
=44/125

(c) Let NA = not approve
P(NA) = 1 - 0.45 = 0.55

P(at most 1 approve) = P(all NA) + 3 P(A.NA.NA)
= 0.55 * 0.55 * 0.55 + 3 (0.45 * 0.55 * 0.55)
= 2299/4000

O lvl E Maths: Probability

Given that
X = { x : x is an integer and 5 ≤ x ≤ 10 }
Y = { y : y is a multiple of 3 and 1 ≤ y ≤ 10 }

An element is chosen at random from each set.
Write down the total number of possible outcomes.

Using a probability diagram or otherwise, find the probability that:
(i) the product xy is a perfect square.
(ii) the product xy is an odd number.
(iii) the product xy > 20.

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Answer:

X = { x : 5, 6, 7, 8, 9, 10}
Y = { y : 3, 6, 9}
Total number of outcomes = 6 numbers * 3 numbers = 18

(i) Perfect squares = 6² and 9²
so, probability that product xy is a perfect square = 2/18 = 1/9

(ii) xy is an odd number means both x and y must be odd numbers
Number of odd numbers in x = 3
Number of odd numbers in y = 3
so, probability that product xy is an odd number = 3 * 3 / 18 = 1/2

(iii) Number of terms of xy < 20 : 5 * 3, 6 * 3
===> only 2 terms.

So, 16 terms are greater than 20
Probability of product xy > 20 = 16/18 = 8/9

O lvl E Maths: Probability

Two bags each contains 91 balls. In each bag, there are 3 red balls, 1 white and the rest blue.

(i) One ball is drawn from the first bag. Find the probability that it is not blue.

(ii) One ball is drawn from the second bag. Find the probability that it is blue.

(iii) One ball is drawn from each bag. Find the probability that the two balls are
(a) both blue,
(b) each of a different colour.

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Answer:

In each bag, there are 3 red balls, 1 white and 87 blue.

(i) P(ball is not blue) = 4/91

(ii) P(ball is blue) = 87/91

(iii)
(a) P(both balls are blue) = 87/91 * 87/91
= 7569/8281

(b) P(both balls same colour)
= P(both red) + P(both white) + P(both blue)
= 3/91 * 3/91 + 1/91 * 1/91 + 7569/91
= 7579/8281

P(both balls are different colour) = 1 - P(both balls same colour)
= 1 - 7579/8281
= 702/8281

O lvl E Maths: Probability

A student attends either a Mathematics class or an Additional mathematics class everyday. The probability of him bringing the wrong textbook is 3/5 if he attends the Mathematics class and 3/10 if he attends the Additional Mathematics class.

(a) If he attends Mathematics class on three consecutive days, find the probability that he will bring the wrong textbook
(i) on all the three days
(ii) on just one of the three days.

(b) If he is equally likely to attend the Mathematics class or the Additional Mathematics class, find the probability that he will bring the correct textbook to class on any given day.

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Answer:

(a)
(i) P(wrong textbook on all three days)
= P(1st day wrong textbook) * P(2nd day wrong textbook) * P(3rd day wrong textbook)
= 3/5 * 3/5 * 3/5
= 27/125

(ii) P(just one of three days wrong textbook)
= P(1st day wrong, 2nd, 3rd day correct) + P(2nd day wrong, 1st, 3rd day correct) + P(3rd day wrong, 1st, 2nd day correct)
= 3/5 * 2/5 * 2/5 + 2/5 * 3/5 * 2/5 + 2/5 * 2/5 * 3/5
= 36/125

(b) P(correct textbook for Mathematics or for Additional Mathematics)
= P(attend Mathematics and bring correct txtbook for Mathematics) + P(attend Additional Mathematics and bring correct txtbook for Additional Mathematics)
= 1/2 * 2/5 + 1/2 * 7/10
= 11/20

O lvl E Maths: Probability

Each member of a class of 30 boys supports one and only one of the three football teams; 13 boys support city, 10 support Rovers, and 7 support United.

(i) If a boy is to be chosen at random, what is the probability that he will support City?
(ii) If 2 boys are to be chosen at random, what is the probability that they will both support City?

Find the probability that
(iii) two boys chosen at random will support the same team.
(iv) two boys chosen will support different teams.

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Answer:

(i) 13/30 (13 boys out of 30 boys)

(ii) P(both boys support City)
= P(1st boy support City) * P(2nd boy support City)
= 13/30 * 12/29
= 26/145

(iii) P(both boys support the same team)
= P(both boys support City) + P(both boys support Rovers) + P(both boys support United)
= 26/145 + 10/30 * 9/29 + 7/30 * 6/29
= 26/145 + 3/29 + 7/145
= 48/145

(iv) P(two boys chosen will support different teams)
= 1 - P(both boys support the same team)
= 1 - 48/145
= 97/145

O lvl E Maths: Probability

A certain brand of boxes of matches is advertised as having "average contents 50 matches". The probability that a box chosen at random will contain exactly 50 matches is 5/8.

(a) Calculate the probability that a box of matches chosen at random will not contain exactly 50 matches.


The probability that a box chosen at random will contain more than 50 matches is twice the probability that it will contain less than 50 matches.

(b) Calculate the probability that
(i) one box chosen at random will contain at least 50 matches.
(ii) of 2 boxes chosen at random, at least one will contain less than 50 matches.

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Answer:

(a) P(a box of matches chosen at random will not contain exactly 50 matches)
= 1 - 5/8
= 3/8


(b)
The probability that a box chosen at random will contain more than 50 matches is twice the probability that it will contain less than 50 matches
====> P(a box chosen at random will contain more than 50 matches) = 2/8
====> P(a box chosen at random will contain less than 50 matches) = 1/8

(i) P(one box chosen at random will contain at least 50 matches)
= P(a box chosen at random will contain more than 50 matches) + P((a box chosen at random will contain exactly 50 matches)
= 2/8 + 5/8
= 7/8

(ii) P(at least one will contain less than 50 matches)
= 1 - P(both boxes contain at least 50 matches)
= 1 - 7/8 * 7/8
= 15/64

O lvl E Maths: Probability

Four bags, A, B, C and D each contains red discs and blue discs which are identical except for the colour.

Bag A contains 8 red discs and 2 blue discs.
Bag B contains 6 red discs and 4 blue discs.
Bag C contains 3 red discs and 5 blue discs.
Bag D contains 1 red discs and 7 blue discs.

(i) A disc is taken at random from A. Find, as a fraction, the probability that the disc is blue.
(ii) A disc is taken at random from B and another disc at random from C. Find, as a fraction, the probability that both discs are blue.
(iii) Two discs are taken at random from D. Find the probability that at least one of the discs is blue.

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Answer:

(i) P(Disc is blue)
= 2/(2 + 8)
= 2/10
=1/5

(ii) P(both discs are blue)
= P(disc from B is blue and disc from C is blue)
= 4/(6+4) * 5/(3+5)
= 4/10 * 5/8
= 1/4

(iii) P(at least one is blue)
= P(B,R) + P(R,B) + P(B,B)
= 7/8 * 1/7 + 1/8 * 7/7 + 7/8 * 6/7
= 1/8 + 1/8 + 6/8
= 1

O lvl E Maths: Probability

(a) An ordinary six-sided die numbered 1 to 6 is thrown. Write down the probability that the number shown on the die is a prime number.

A six-sided die numbered 1 to 6 and an eight sided die numbered 3 to 10 are thrown together. Giving each answer as a fraction in its lowest terms, find the probability that
(i) the sum of the numbers is 10,
(ii) the two numbers are not equal.

(b) A pupil has difficulty in waking up for school, and so, to wake himself, he sets 3 alarms to go off at the same time, as the noise from at least 2 alarms is necessary to wake him. Each alarm goes off independently.
The probability that each alarm goes off is 0.7, 0.8 and 0.9 respectively. Find the probability that
(i) all 3 go off,
(ii) the pupil is awakened.

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Answer:

(a) Prime numbers are 2, 3, 5
Hence, probability that it is a prime number = 3/6 = 1/2

(i) P(sum of numbers is 10)
= P(1,9) + P(2,8) + P(3,7) + P(4,6) + P(5,5) + P(6,4) => Summing all the possible combinations
= 1/6 * 1/8 * 6 times
= 1/8

(ii) P(two numbers are not equal)
= 1 - P(two numbers are equal)
= 1 - {P(3,3) + P(4,4) + P(5,5) + P(6,6)}
= 1 - {1/6 * 1/8 * 4 times}
= 1 - 1/12
= 11/12


(b)
(i) Probability that all 3 go off
= 7/10 * 8/10 * 9/10
= 63/125

(ii) P(Student is awakened)
= P(at least 2 alarms goes off)
= P(1,2, 3 don't go off) + P(2,3, 1 don't go off) + P(1,3, 2 don't go off) + P(all 3 go off) ==> P(which alarm goes off)
= 0.7 * 0.8 * 0.1 + 0.8 * 0.9 * 0.3 + 0.7 * 0.9 * 0.2 + 63/125
= 0.398 + 63/125
= 451/500

=

O lvl E Maths: Probability

RGS Practice Question

Two dice are thrown. One is biased so that the probability of throwing a 6 is 1/16 and the probabilities of throwing the other five numbers are all equal. The other die is unbiased. Calculate, as a fraction in its lowest term, the probability that
(i) both numbers thrown are not 6,
(ii) at least one of the two numbers thrown is a 6,
(iii) exactly one of the numbers thrown is a 4.

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Answer:

(i) P(both numbers thrown are not 6)
= P(1st dice not 6 and 2nd dice not 6)
= 15/16 * 5/6
= 25/32

(ii) P(at least one of the two numbers thrown is a 6)
= 1 - P(both numbers thrown are not 6)
= 1 - 25/32
= 7/32

(iii) For the biased dice, the probability of getting 1, 2, 3, 4 or 5 is (1 - 1/16)/5 = 3/16

P(exactly one of the numbers thrown is a 4)
= P(1st dice is 4, and 2nd dice is not 4) + P(1st dice is not 4, and 2nd dice is 4)
= 3/16 * 5/6 + 13/16 * 1/6
= 5/32 + 13/96
= 7/24

O lvl E Maths: Probability

Nov 2002 P1 Q16

(1) Two unbiased dice are thrown.

Find the probability that they
(i) show the same number
(ii) show different number

(2) Three unbiased dice are thrown.

Find the probability that
(i) they all show different numbers
(ii) at least two show the same number

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Answer:

(1)
(i) P(both shows same number) = 1/6 * 1/6 * 6 (6 different numbers) = 1/6

(ii) P(both shows different number) = 1 - both show same number = 5/6

(2)
(i) P(all different numbers) = P(first 2 dice are different and 3rd dice is also different number)

= P(first 2 dice different) * P(3rd dice different number from first 2 dice)
= 5/6 * 4/6 (4 numbers out of 6 left since the first 2 dice have already taken/used up 2 numbers)
= 5/9

(ii) P (at least two show same number) = 1 - P(all different numbers) = 4/9

O lvl E Maths: Probability

Dunman High Prelims 2008 P1 Q9

A small box containing chocolates of which 3 contains hazelnuts, 7 contains almonds and 2 contain sultanas. Two chocolates are picked at random, one at a time, without replacement. Find the probability that
(a) both contain hazelnuts,
(b) one contains almonds and one contains hazelnuts,
(c) both are without sultanas.

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Answer:

(a) Probability that both contain hazelnuts
= probability of first chocolate containing hazelnuts and probability of second chocolate containing hazelnuts
= 3/12 * 2/11
= 1/22

(b) Probability that one contains almonds and one contains hazelnuts
= probability of first chocolate containing almonds and probability of second chocolate containing hazelnuts OR probability of first chocolate containing hazelnuts and probability of second chocolate containing almonds
= 7/12 * 3/11 + 3/12 * 7/11
= 7/44 + 7/44
= 7/22

(c) Probability that both are without sultanas
= probability of first chocolate NOT containing sultanas and probability of second chocolate NOT containing sultanas
= 10/12 * 9/11
= 15/22