Sample Space Diagrams

Imagine one row for each filling.

Keep the tens digit fixed while trying every possible units digit.

An outcome must describe both the coin and the spinner.

Use one grid row for each result on A.

Read the row and column headings of the missing cell.

Choose the hundreds digit first, then arrange the remaining two digits in both orders.

1. Starting with 2 gives 247 and 274. Starting with 4 gives 427 and 472. Starting with 7 gives 724 and 742.
2. The complete list is 247, 274, 427, 472, 724, 742. No digit is reused within a number.

HH has two heads, so it does not qualify.

1. Of HH, HT, TH and TT, only HT and TH have exactly one head.
2. The probability is \frac24=\frac12.

A grid cell adds its row heading and column heading. Each pair of spins is equally likely.

Part (a)

1. x is in row 2, column 2. Therefore x=2+2=4.

Part (b)

1. The completed grid has nine equally likely cells. Three show 4: (1,3), (2,2) and (3,1).
2. The probability is \frac39=\frac13.

Do not include a total of 4.

1. The successful pairs are (2,3), (3,2) and (3,3), with totals 5, 5 and 6.
2. There are 9 equally likely pairs, so the probability is \frac39=\frac13.

The second spinner cannot show 3 or 4.

1. There are 4\times2=8 equally likely pairs. Only (1,1) and (2,2) match.
2. The probability is \frac28=\frac14.

Use the correct spinner value in each row and column. Count all equally likely pairs when finding a probability.

Part (a)

1. The blank in row 2, column 3 is 2+3=5. The blank in row 3, column 4 is 3+4=7.
2. The two missing totals are 5 and 7.

Part (b)

1. Total 6 occurs for (2,4) and (3,3). No other pair of available numbers gives 6.
2. There are 3\times4=12 equally likely pairs, so the probability is \frac2{12}=\frac16.

Use a product grid and count even entries.

1. A product is even when at least one result is 2. The pairs are (1,2), (2,1), (2,2), (2,3), (3,2).
2. Five of the nine equally likely pairs qualify, so the probability is \frac59.

Distinguish possible outcomes from equally likely outcomes.

1. The grid correctly lists four possible outcomes, but the weighted coin does not give all pairs the same chance. Four possible outcomes alone does not make each probability \frac14.

List one die’s result in order, then find the partner needed for 9.

1. The successful ordered pairs are (3,6), (4,5), (5,4) and (6,3).
2. There are 6\times6=36 equally likely pairs, so the probability is \frac4{36}=\frac19.

“At least 10” includes 10, 11 and 12.

1. Total 10: (4,6), (5,5), (6,4). Total 11: (5,6), (6,5). Total 12: (6,6).
2. There are 3+2+1=6 successful pairs out of 36. The probability is \frac6{36}=\frac16.

For each number on spinner A, multiply by every number on spinner B.

1. For A = 1, no product is greater than 10. For A = 2, only B = 6 succeeds. For A = 3, B = 4 or 6 succeeds.
2. The pairs (2,6), (3,4), (3,6) give 3 successes out of 9, so the probability is \frac13.

List each winning condition and check whether any pair appears twice.

1. Total 5 gives (1,4), (2,3), (3,2), (4,1): four pairs. Equal numbers give (1,1), (2,2), (3,3), (4,4): four more.
2. No pair appears in both lists. There are 8 winning pairs out of 16, giving \frac8{16}=\frac12.

A difference of 1 can happen in either order.

1. The adjacent pairs in increasing order are (1,2), (2,3), (3,4), (4,5), (5,6). Each can also occur in reverse order.
2. That gives 5\times2=10 pairs out of 36. The probability is \frac{10}{36}=\frac5{18}.

Use the same 16 pairs but apply a different calculation for each event.

1. Total 5: (1,4), (2,3), (3,2), (4,1), so probability \frac4{16}.
2. Product 6: (2,3), (3,2), so probability \frac2{16}.
3. The difference is \frac4{16}-\frac2{16}=\frac2{16}=\frac18.

Repeated labels still represent separate equally likely sectors.

1. Keep the two sectors labelled 1 on A as separate rows. Each can match B’s 1, giving two successful sector pairs. A’s 2 can match either sector labelled 2 on B, giving two more.
2. All 3\times3=9 sector pairs are equally likely. The probability is \frac49.