Unit notes and practice questions
Everything here is for unit 8 of the Grade 11 Mathematics textbook: what the chapter covers, the short notes, practice questions with their answers, and the lesson video where one exists.
10 multiple-choice questions written against this chapter, each with its answer and the reason. Try one before you open the answer.
Question 1
A fair die is rolled once. How many outcomes are in the event of getting an odd number?
Answer: B 3
Why
The event of getting an odd number is {1, 3, 5}. This event has 3 outcomes. The sample space {1, 2, 3, 4, 5, 6} has 6 outcomes, but the question asks only about the odd numbers.
Question 2
A fair coin is tossed three times. How many outcomes are in the sample space?
Answer: C 8
Why
The textbook lists the sample space as HHH, HHT, HTH, THH, TTH, THT, HTT and TTT. That is 8 outcomes. Each toss has 2 results, so 2 x 2 x 2 = 8.
Question 3
A coin is tossed and a die is rolled at the same time. How many possible outcomes are there for this sequence of events?
Answer: C 12
Why
A coin has 2 outcomes (H and T) and a die has 6 outcomes (1 to 6). By the multiplication principle, 2 x 6 = 12 possible outcomes.
Question 4
When a die is rolled, the event of getting an odd number is which type of event?
Answer: B Compound event
Why
A simple event has only one sample element. A compound event has more than one. The event of getting an odd number is {1, 3, 5}, which has three elements, so it is a compound event.
Question 5
Tigist is choosing an outfit. She has 5 different pairs of pants, 4 different skirts and 12 different T-shirts. She wears either pants or a skirt, and always wears a T-shirt. How many outfits are possible?
Answer: D 108
Why
Pants with a T-shirt give 5 x 12 = 60 outfits. Skirts with a T-shirt give 4 x 12 = 48 outfits. Because she cannot wear pants and a skirt together, we add: 60 + 48 = 108.
Question 6
In how many ways can five people be seated at a round table?
Answer: B 24
Why
For circular arrangements, the number of permutations of n elements in a circle is (n - 1)!. Here n = 5, so (5 - 1)! = 4! = 4 x 3 x 2 x 1 = 24.
Question 7
Compute P(7, 4).
Answer: C 840
Why
P(n, r) = n! / (n - r)!. So P(7, 4) = 7! / 3! = 7 x 6 x 5 x 4 = 840.
Question 8
Using Pascal's triangle, what is the coefficient of a2b2 in the expansion of (a + b)4?
Answer: B 6
Why
The row for n = 4 in Pascal's triangle is 1, 4, 6, 4, 1. The expansion is a4 + 4a3b + 6a2b2 + 4ab3 + b4. The coefficient of a2b2 is 6.
Question 9
How many different permutations can be made from the letters of the word ADDIS if the arrangement must start with the letter S?
Answer: A 12
Why
If S is fixed as the first letter, four letters remain: A, D, D and I. There are 2 D's, so the number of arrangements is 4! / 2! = 24 / 2 = 12. The textbook lists these 12 arrangements.
Question 10
In an HIV/AIDS club there are 7 women and 5 men. A committee of 3 women and 2 men is to be chosen. How many different committees are possible?
Answer: D 350
Why
The 3 women can be chosen from 7 in C(7, 3) = 35 ways. The 2 men can be chosen from 5 in C(5, 2) = 10 ways. Because the women and men are chosen separately, we multiply: 35 x 10 = 350.
The app holds 12 questions on this unit, marked as you go. These are the first 10.
A lesson on this chapter from YouTube. It loads only when you press play.
Binomial Theorem (part 2) · Watch on YouTube
Unit 8 runs from page 413 to page 482 of the Ministry of Education edition.
Ministry of Education edition, English, 494 pages
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