Question

determine the probability of each
individual event. then, determine the
probability of each compound event.
show your calculations.

sample problem

the shell game consists of placing three
opaque cups, representing shells, upside
down on a table and hiding a ball under one
of the cups. a player, who has not seen
where the ball is hidden, has to choose one
of the cups. if the ball is hidden under it, the...

enter the answer in each space provided. use numbers instead of
words.

p(any block 1st) =

p(w 2nd) =

Explanation:

To solve the probability problems, we first need to determine the total number of blocks and then the number of favorable outcomes for each event.

Step 1: Count the total number of blocks

Let's count the blocks row by row:

• First row: 4 (A, C, D, B) + 4 (Y, Y, Z, X) = 8
• Second row: 4 (A, B, B, A) + 4 (Y, W, W, Z) = 8
• Third row: 4 (B, B, C, D) + 4 (Z, Y, X, W) = 8

Wait, actually, looking at the grid, let's count each cell:

First row (left 4, right 4): 4 + 4 = 8
Second row (left 4, right 4): 4 + 4 = 8
Third row (left 4, right 4): 4 + 4 = 8

Total blocks: \( 8 + 8 + 8 = 24 \)? Wait, no, let's count each block individually:

Left section (3 rows, 4 columns): \( 3 \times 4 = 12 \)
Right section (3 rows, 4 columns): \( 3 \times 4 = 12 \)
Total blocks: \( 12 + 12 = 24 \).

Part 1: \( P(\text{any block 1st}) \)

Wait, "any block 1st" – maybe it's a typo or misstatement. Wait, maybe "any block (1st position? No, maybe "any block" – but actually, if we consider "any block" as in selecting any block, the probability would be 1, but that doesn't make sense. Wait, maybe it's "any block labeled A" or "any block in the first row"? Wait, the sample problem is about the shell game, but the actual problem here is about blocks. Let's re-examine the blocks:

Let's list all blocks:

Left section (3 rows, 4 columns):
Row 1: A, C, D, B
Row 2: A, B, B, A
Row 3: B, B, C, D

Right section (3 rows, 4 columns):
Row 1: Y, Y, Z, X
Row 2: Y, W, W, Z
Row 3: Z, Y, X, W

Wait, maybe the problem is:

1. \( P(\text{any block (1st)}) \) – maybe "any block in the first row"? Let's check the first row:

First row has 8 blocks (4 left, 4 right). Total blocks: 24. So \( P(\text{first row block}) = \frac{8}{24} = \frac{1}{3} \). But that seems off. Wait, maybe "any block labeled with a letter (but all are letters)". Wait, maybe the first problem is \( P(\text{any block}) \), which is 1, but that's not helpful. Wait, maybe it's a translation error or typo. Alternatively, maybe "any block with a vowel" (A, E, I, O, U – but here we have A, E? Wait, left section row 2: A, B, B, A – so A is a vowel. Let's check:

Left section:
Row 1: A (vowel), C (consonant), D (consonant), B (consonant)
Row 2: A (vowel), B (consonant), B (consonant), A (vowel)
Row 3: B (consonant), B (consonant), C (consonant), D (consonant)

So vowels in left: A, A, A → 3? Wait, row 1: 1 A, row 2: 2 A's, row 3: 0. So total vowels in left: 3. Right section: Y, Y, Z, X, Y, W, W, Z, Z, Y, X, W – all consonants (Y, Z, X, W are consonants). So total vowels: 3. Total blocks: 24. So \( P(\text{vowel}) = \frac{3}{24} = \frac{1}{8} \). But this is guesswork. Wait, maybe the first problem is \( P(\text{any block}) \), which is 1, but that's not possible. Alternatively, maybe "any block in the first column". First column:

Left first column: A (row1), A (row2), B (row3)
Right first column: Y (row1), Y (row2), Z (row3)
Total first column blocks: 3 + 3 = 6. Total blocks: 24. So \( P(\text{first column}) = \frac{6}{24} = \frac{1}{4} \). Still not sure.

Wait, maybe the second problem is \( P(\text{W 2nd}) \) – "W in the second position"? Wait, "2nd" could mean second row or second column. Let's check W's:

Right section:
Row 2: W, W
Row 3: W
So total W's: 3 (row2: 2, row3:1). Wait, row2 right: Y, W, W, Z → 2 W's. Row3 right: Z, Y, X, W → 1 W. So total W's: 3. Total blocks: 24. So \( P(\text{W}) = \frac{3}{24} = \frac{1}{8} \). But maybe "W in the second row". Second row has 8 blocks (4 left, 4 right). W's in second row: 2 (right section row2: W, W). So \( P(\text{W in second row}) = \frac{2}{8} = \frac{1}{4} \…

Step1: Count total blocks

Total blocks = 3 rows × 8 columns = 24.

Step2: Probability of any block in first row

Number of first row blocks = 8.
\( P(\text{first row}) = \frac{8}{24} = \frac{1}{3} \).

Step3: Probability of W

Number of W's = 3 (row2: 2, row3:1).
\( P(\text{W}) = \frac{3}{24} = \frac{1}{8} \).

Answer:

\( P(\text{any block 1st}) = \frac{1}{3} \) (or 0.333)
\( P(\text{W 2nd}) = \frac{1}{8} \) (or 0.125)