CAIE Mathematics (9709)

2023 Oct/Nov
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Questions & Answers

Q5a(i).
Short Answer 2 marks

The heights of the members of a club are normally distributed with mean 166 cm and standard deviation 10 cm. Find the probability that a randomly chosen member of the club has height less than 170 cm.

Model Answer:
P(X < 170) = P(Z < (170-166)/10) = P(Z < 0.4) = 0.6554
Explanation: Standardise using the normal distribution and read the cumulative probability from the standard normal table.
Q5a(ii).
Short Answer 3 marks

The heights of the members of a club are normally distributed with mean 166 cm and standard deviation 10 cm. Given that 40% of the members have heights greater than h cm, find the value of h correct to 2 decimal places.

Model Answer:
P(X > h) = 0.40, so P(X < h) = 0.60. Hence (h-166)/10 = 0.2533, giving h = 168.53 cm.
Explanation: The 60th percentile of the standard normal distribution is approximately 0.2533.
Q5b.
Short Answer 3 marks

The random variable X is normally distributed with mean and standard deviation 3. Given that 3 = 2/3 -, find the probability that a randomly chosen value of X is positive.

Model Answer:
P(X > 0) = P(Z > (0-μ)/3). Using the given relation, the standardised value is 2/3, so P(X > 0) = P(Z > -2/3) = 0.7475.
Explanation: Use the normal distribution and the symmetry of the standard normal curve.
Q6a.
Structured 3 marks

Freddie has two bags of marbles. Bag X contains 7 red marbles and 3 blue marbles. Bag Y contains 4 red marbles and 1 blue marble. Freddie chooses one of the bags at random. A marble is removed at random from that bag and not replaced. A new red marble is now added to each bag. A second marble is then removed at random from the same bag that the first marble had been removed from. Draw a tree diagram to represent this information, showing the probability on each of the branches.

Model Answer:
First choose bag X or Y with probability 1/2 each. If bag X is chosen: first draw red 7/10 or blue 3/10; after adding a red marble, second draw probabilities depend on the first draw. If first was red, bag X becomes 7 red, 3 blue again, so second draw red 7/10, blue 3/10. If first was blue, bag X becomes 8 red, 2 blue, so second draw red 8/10, blue 2/10. If bag Y is chosen: first draw red 4/5 or blue 1/5; after adding a red marble, if first was red then second draw red 4/5, blue 1/5; if first was blue then second draw red 5/5, blue 0/5.
Explanation: The tree branches follow from the bag composition after the first marble is removed and one red marble is added to each bag.
Q6b.
Short Answer 4 marks

Freddie has two bags of marbles. Bag X contains 7 red marbles and 3 blue marbles. Bag Y contains 4 red marbles and 1 blue marble. Freddie chooses one of the bags at random. A marble is removed at random from that bag and not replaced. A new red marble is now added to each bag. A second marble is then removed at random from the same bag that the first marble had been removed from. Find the probability that both of the marbles removed from the bag are the same colour.

Model Answer:
P(same colour) = 1/2[(7/10)(7/10) + (3/10)(2/10)] + 1/2[(4/5)(4/5) + (1/5)(0/5)] = 1/2(49/100 + 6/100) + 1/2(16/25) = 55/200 + 32/100 = 23/40.
Explanation: Add the probabilities of RR and BB for each bag, then average over the two equally likely bags.
Q6c.
Short Answer 2 marks

Freddie has two bags of marbles. Bag X contains 7 red marbles and 3 blue marbles. Bag Y contains 4 red marbles and 1 blue marble. Freddie chooses one of the bags at random. A marble is removed at random from that bag and not replaced. A new red marble is now added to each bag. A second marble is then removed at random from the same bag that the first marble had been removed from. Find the probability that bag Y is chosen given that the marbles removed are not both the same colour.

Model Answer:
P(Y | not same) = P(Y and not same) / P(not same) = [1/2((4/5)(1/5) + (1/5)(1))] / [1 - 23/40] = (1/2)(8/25)/17/40 = 16/85.
Explanation: Use conditional probability with the probability of not both the same colour from part (b).
Q7a.
Short Answer 3 marks

Find the number of different arrangements of the 9 letters in the word ANDROMEDA in which no consonant is next to another consonant. (The letters D, M, N and R are consonants and the letters A, E and O are not consonants.)

Model Answer:
There are 4 consonants and 5 vowels, with A repeated twice. Arrange the vowels in 5!/2! = 60 ways. There are 6 gaps around the vowels, and choose 4 of them for the consonants: C(6,4) = 15. Arrange the 4 consonants in 4! = 24 ways. Total = 60 × 15 × 24 = 21600.
Explanation: Place consonants in separate gaps between vowels to ensure no two consonants are adjacent.
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