← Back to the Counting Probability worksheet
Last updated: October 22, 2026 · Written by the Math Made Easy math teachers
Standards: S-CP.B.8, S-CP.B.9, S-MD.A.1, S-MD.A.2 · Level: Grade 12 Precalculus
Problem 1 Sandwich Shop Combinations
Multiply the choices: 5 × 4 × 3 = 60.
Answer: 60 sandwiches
Problem 2 License Plates
26 choices for each letter and 10 for each digit: 26 × 26 × 10 × 10 × 10 = 676,000.
Answer: 676,000 plates
Problem 3 Ordered Picks
Order matters: P(9, 3) = 9 · 8 · 7 = 504.
Answer: 504 ways
Problem 4 Choosing a Group
C(10, 4) = 10 · 9 · 8 · 74 · 3 · 2 · 1 = 5,04024 = 210.
Answer: 210 groups
Problem 5 Order or Not?
(a) The two offices are different, so order matters: P(12, 2) = 12 · 11 = 132.
(b) Two delegates form an unordered pair: C(12, 2) = 12 · 112 = 66.
Answer: (a) 132 by permutation; (b) 66 by combination
Problem 6 Candy Jar Probability
The jar holds 7 + 9 + 4 = 20 candies.
(a) P(cherry) = 920.
(b) Not lemon means 9 + 4 = 13 candies, so 1320.
Answer: (a) 920; (b) 1320
Problem 7 Books on a Shelf
(a) 6! = 720.
(b) Glue the two books into one block: 5 items can be arranged in 5! = 120 ways, and the two books can swap inside the block in 2 ways. So 120 × 2 = 240.
Answer: (a) 720; (b) 240
Problem 8 Starting Lineups
C(12, 5) = 12 · 11 · 10 · 9 · 85! = 95,040120 = 792.
Answer: 792 ways
Problem 9 Arranging Letters
BANANAS has 7 letters: B once, A three times, N twice, S once.
Divide out the repeats: 7!3! · 2! = 5,04012 = 420.
Answer: 420 arrangements
Problem 10 All Juniors
Favorable: C(5, 4) = 5. Total: C(12, 4) = 495.
P = 5495 = 199 ≈ 0.0101.
Answer: 199
Problem 11 Two Blue Without Replacement
P(first blue) = 712. After removing one blue, 6 blue remain among 11, so P(second blue | first blue) = 611.
P(both blue) = 712 · 611 = 42132 = 722.
Answer: 722
Problem 12 Red Then Blue
P(red first) = 512. Then 7 blue remain among 11, so P(blue second | red first) = 711.
P = 512 · 711 = 35132 ≈ 0.265.
Answer: 35132
Problem 13 With or Without Replacement
With replacement the draws are independent: 712 · 712 = 49144 ≈ 0.340.
Without replacement the answer is 722 ≈ 0.318.
Without replacement the probability is smaller, because removing a blue ball leaves fewer blue balls.
Answer: 49144 ≈ 0.340 compared with 722 ≈ 0.318
Problem 14 Expected Value from a Table
The probabilities add to 0.1 + 0.4 + 0.3 + 0.2 = 1. E(X) = 0(0.1) + 1(0.4) + 2(0.3) + 3(0.2) = 0 + 0.4 + 0.6 + 0.6 = 1.6.
Answer: 1.6
Problem 15 Carnival Spinner
Expected winnings: 10(0.1) + 3(0.3) + 0(0.6) = 1 + 0.9 = $1.90.
Net gain = 1.90 − 2.00 = −$0.10.
On average the player loses 10 cents per spin.
Answer: $1.90 expected winnings; expected net −$0.10
Problem 16 Password Count
The letters are chosen without repeats: P(26, 4) = 26 · 25 · 24 · 23 = 358,800. The digits give 10 × 10 = 100.
Total = 358,800 × 100 = 35,880,000.
Answer: 35,880,000 passwords
Problem 17 Committee with Boys
Use the complement. Total committees: C(25, 5) = 53,130.
No boys: C(14, 5) = 2,002. Exactly one boy: C(11, 1) · C(14, 4) = 11 × 1,001 = 11,011.
P(at most 1 boy) = 2,002 + 11,01153,130 = 13,01353,130.
P(at least 2 boys) = 1 − 13,01353,130 = 40,11753,130 ≈ 0.755.
Answer: about 0.755
Problem 18 Light Bulbs
P(all work) = 812 · 711 · 610 = 3361,320 = 1455 ≈ 0.255.
Check with combinations: C(8, 3)C(12, 3) = 56220 = 1455.
Answer: 1455
Problem 19 Raffle Expected Value
Total prize money: 400 + 3(100) + 10(20) = 400 + 300 + 200 = $900.
Expected winnings per ticket = 900500 = $1.80. Expected net gain = 1.80 − 4.00 = −$2.20.
Answer: $1.80 expected winnings; expected net −$2.20
Problem 20 Defective Phones
Total pairs: C(12, 2) = 66.
(a) P(0) = C(9, 2)66 = 3666 = 611. P(1) = 3 · 966 = 2766 = 922. P(2) = C(3, 2)66 = 366 = 122. The three add to 1.
(b) E(X) = 0 · 611 + 1 · 922 + 2 · 122 = 1122 = 12.
Answer: P(0) = 611, P(1) = 922, P(2) = 122; E(X) = 12
Mis à jour le