Tree Diagrams (B) Worksheet | PDF Printable Statistics ... - Free Printable

Here are the step-by-step solutions for the problems on the worksheet.

Problem 1

Scenario: A box has 12 red cubes and 6 green cubes. Total = 18 cubes. We pick one, keep it out (no replacement), and pick another.

Completing the Tree Diagram Probabilities:
* First Choice:
* Red (R): $12/18$ which simplifies to $2/3$.
* Green (G): $6/18$ which simplifies to $1/3$.
* Second Choice (if first was Red): There are now 17 cubes left (11 Red, 6 Green).
* Red (R): $11/17$
* Green (G): $6/17$
* Second Choice (if first was Green): There are now 17 cubes left (12 Red, 5 Green).
* Red (R): $12/17$
* Green (G): $5/17$

Calculations:

a. Probability of choosing Two red cubes (R, R)
Multiply the probabilities along the top branch:
$$ \frac{2}{3} \times \frac{11}{17} = \frac{22}{51} $$

b. Probability of choosing At least one green cube
"At least one green" means we get either (Red, Green), (Green, Red), or (Green, Green). It is often easier to calculate the opposite event ("No green," which means "Two Reds") and subtract that from 1.
$$ P(\text{At least one G}) = 1 - P(\text{Two Reds}) $$
$$ 1 - \frac{22}{51} = \frac{29}{51} $$

*(Check by adding individual branches: RG + GR + GG)*
* $RG = \frac{2}{3} \times \frac{6}{17} = \frac{12}{51}$
* $GR = \frac{1}{3} \times \frac{12}{17} = \frac{12}{51}$
* $GG = \frac{1}{3} \times \frac{5}{17} = \frac{5}{51}$
* Sum: $\frac{12+12+5}{51} = \frac{29}{51}$. The answer matches.

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Problem 2

Scenario: A class has 20 boys and 10 girls. Total = 30 students. Two students are chosen at random without replacement.

Completing the Tree Diagram Probabilities:
* First Choice:
* Boy (B): $20/30$ simplifies to $2/3$.
* Girl (G): $10/30$ simplifies to $1/3$.
* Second Choice (if first was Boy): 29 students left (19 Boys, 10 Girls).
* Boy (B): $19/29$
* Girl (G): $10/29$
* Second Choice (if first was Girl): 29 students left (20 Boys, 9 Girls).
* Boy (B): $20/29$
* Girl (G): $9/29$

Calculations:

a. If the first student chosen is a girl, what is the probability that the second student is also a girl?
This asks for the conditional probability on the bottom branch. After picking one girl, there are 9 girls left out of 29 total students.
Answer: $9/29$

b. What is the probability that a girl and boy are chosen?
This can happen in two ways: (Girl then Boy) OR (Boy then Girl).
* Path 1 (G, B): $\frac{1}{3} \times \frac{20}{29} = \frac{20}{87}$
* Path 2 (B, G): $\frac{2}{3} \times \frac{10}{29} = \frac{20}{87}$
* Total Probability: $\frac{20}{87} + \frac{20}{87} = \frac{40}{87}$

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Problem 3

Scenario: A box has 200 nails. 5 are too small (S), so 195 are normal size (S'). Three nails are selected.

a. If the first nail chosen is too small, what is the probability that the second nail is also too small?
If the first nail was Small (S), there are now 4 Small nails left and 199 total nails left.
Answer: $4/199$

b. What is the probability that All three nails are too small (S, S, S)?
We multiply the probabilities for picking Small three times in a row.
1st pick S: $5/200$
2nd pick S: $4/199$
3rd pick S: $3/198$

Calculation:
$$ \frac{5}{200} \times \frac{4}{199} \times \frac{3}{198} $$
Simplify fractions first:
$\frac{5}{200} = \frac{1}{40}$
$\frac{3}{198} = \frac{1}{66}$

$$ \frac{1}{40} \times \frac{4}{199} \times \frac{1}{66} $$
Cancel the 4 and 40 ($\frac{4}{40} = \frac{1}{10}$):
$$ \frac{1}{10} \times \frac{1}{199} \times \frac{1}{66} = \frac{1}{10 \times 199 \times 66} $$
$$ 10 \times 66 = 660 $$
$$ 660 \times 199 = 131,340 $$
Answer: $1/131,340$

c. What is the probability that Only the second nail is too small?
This specific order is: Normal (S'), Small (S), Normal (S').
1st pick S': $195/200$
2nd pick S: $5/199$ (5 small remain, 199 total remain)
3rd pick S': $194/198$ (194 normal remain, 198 total remain)

Calculation:
$$ \frac{195}{200} \times \frac{5}{199} \times \frac{194}{198} $$
Simplify $\frac{195}{200}$ by dividing by 5 $\rightarrow \frac{39}{40}$.
Now we have: $\frac{39}{40} \times \frac{5}{199} \times \frac{194}{198}$

Combine numerators and denominators:
Numerator: $39 \times 5 \times 194 = 37,830$
Denominator: $40 \times 199 \times 198 = 1,576,080$

Fraction: $\frac{37,830}{1,576,080}$
Divide both by 10: $\frac{3,783}{157,608}$
Both are divisible by 3:
$3,783 \div 3 = 1,261$
$157,608 \div 3 = 52,536$

Answer: $1,261 / 52,536$

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Final Answer:

1)
a. $22/51$
b. $29/51$

2)
a. $9/29$
b. $40/87$

3)
a. $4/199$
b. $1/131,340$
c. $1,261/52,536$