Compound Events Worksheet Answer Key – Grade 7

Problem 1 Two coin flips

Sample space: HH, HT, TH, TT. Exactly one head: HT and TH.

P = 2/4 = 1/2.

1/2

Problem 2 Outfit choices

Outfits: red-jeans, red-khaki, blue-jeans, blue-khaki, green-jeans, green-khaki (3 × 2 = 6).

Blue shirt: 2 outcomes out of 6, so P = 2/6 = 1/3.

6 outfits; P(blue shirt) = 1/3

Problem 3 Two spinners

There are 3 × 2 = 6 outcomes. A sum of 4 happens for (2, 2) and (3, 1).

P = 2/6 = 1/3.

1/3

Problem 4 Sum of eight

The 36 outcomes are shown. A sum of 8 appears in 5 cells: (2, 6), (3, 5), (4, 4), (5, 3), (6, 2).

P = 5/36.

5/36

Problem 5 Coin and spinner tree

The four outcomes are HR, HB, TR, TB, and TR is one of them.

P = 1/4.

1/4

Problem 6 Counting a menu

Outcomes: 4 × 3 × 2 = 24.

One particular meal is 1 outcome out of 24.

24 meals; P = 1/24

Problem 7 At least one head

Sample space: HH, HT, TH, TT. Outcomes with at least one head: HH, HT, TH.

P = 3/4.

3/4

Problem 8 Small sums

Sums less than 5: 2 (1 way), 3 (2 ways), 4 (3 ways), which is 1 + 2 + 3 = 6 outcomes.

P = 6/36 = 1/6.

1/6

Problem 9 Coin and four-section spinner

The sample space has 2 × 4 = 8 outcomes. Heads with 3 or 4: H3 and H4.

P = 2/8 = 1/4.

1/4

Problem 10 Three coin flips

Sample space: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT (8 outcomes).

Exactly two heads: HHT, HTH, THH.

P = 3/8.

3/8

Problem 11 Draw with replacement

With replacement, each draw has 5 choices, so there are 5 × 5 = 25 equally likely ordered pairs.

Both red: 3 × 3 = 9 pairs.

P = 9/25.

9/25

Problem 12 Draw without replacement

First draw: 5 choices. Second draw: 4 choices. There are 5 × 4 = 20 ordered pairs.

Both red: 3 × 2 = 6 pairs.

P = 6/20 = 3/10.

3/10

Problem 13 Family of three

The trials with no girl are TTT, which appears 3 times. So 10 − 3 = 7 trials have at least one head.

Experimental: 7/10 = 0.7. Theoretical: 7/8 = 0.875.

The experimental value is lower, which is reasonable with only 10 trials.

0.7 compared with 0.875

Problem 14 Sums of four-section spinners

There are 4 × 4 = 16 outcomes. Sums of at least 7 appear for (3, 4), (4, 3) and (4, 4).

P = 3/16.

3/16

Problem 15 Pizza options

Pizzas: 2 × 3 × 2 = 12.

Thick and large: 2 pizzas (cheese or pepperoni), so 2/12 = 1/6.

Cheese: 2 × 3 = 6 pizzas, so 6/12 = 1/2.

12 pizzas; P(thick and large) = 1/6; P(cheese) = 1/2

Problem 16 Two-digit codes

There are 5 × 5 = 25 codes.

Same digits: 22, 44, 55, 77, 99, so 5/25 = 1/5.

First greater: 4 + 3 + 2 + 1 = 10 codes (42, 52, 54, 72, 74, 75, 92, 94, 95, 97), so 10/25 = 2/5.

1/5 and 2/5

Problem 17 Product game

There are 36 outcomes. List the winning pairs by the first number: (2, 6); (3, 4), (3, 5), (3, 6); (4, 3), (4, 4), (4, 5), (4, 6); (5, 3), (5, 5), (5, 6), (5, 4); (6, 2), (6, 3), (6, 4), (6, 5), (6, 6).

That is 1 + 3 + 4 + 4 + 5 = 17 pairs.

P = 17/36.

17/36

Problem 18 Guessing on a quiz

Trials with exactly two heads: HTH, HHT, THH, HTH, HHT, HHT, which is 6 trials. Experimental: 6/10 = 0.6.

Theoretical: 3 outcomes (HHT, HTH, THH) out of 8, so 3/8 = 0.375.

The difference is 0.225. Ten trials is a small number, so the simulation result can vary from the theoretical value.

0.6 experimental; 3/8 theoretical; the difference comes from chance in a small number of trials

Problem 19 Prime sums

There are 3 × 4 = 12 outcomes. The prime sums are 2, 3, 5 and 7.

Row 1: 2, 3, 4, 5 has 3 primes. Row 2: 3, 4, 5, 6 has 2 primes. Row 3: 4, 5, 6, 7 has 2 primes. Total: 7.

P = 7/12.

7/12

Problem 20 Restaurant combos

Combinations: 3 × 5 × 2 = 30.

Tuna: 3 × 1 × 2 = 6 combinations, so 6/30 = 1/5.

Rye with swiss: 1 × 5 × 1 = 5 combinations, so 5/30 = 1/6.

30 combinations; P(tuna) = 1/5; P(rye and swiss) = 1/6