Probability (upper elem/middle) Word Problems - Free Printable

Let's solve each problem step by step from the worksheet titled "Probability".

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Problem 1:

> A student had to be chosen from Ann’s class. There were 21 boys and 11 girls in total in her class. What is the chance that a randomly selected student will be:
> - (a) a boy?
> - (b) a girl?

Total students = 21 boys + 11 girls = 32 students

(a) Probability of selecting a boy:
$$
P(\text{boy}) = \frac{\text{number of boys}}{\text{total students}} = \frac{21}{32}
$$

(b) Probability of selecting a girl:
$$
P(\text{girl}) = \frac{11}{32}
$$

✔ Answer:
- (a) $ \frac{21}{32} $
- (b) $ \frac{11}{32} $

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Problem 2:

> Thomas had a bag with 10 marbles: 12 blue, 7 red, 5 white, and 4 black. Wait — this doesn't add up! Let's check:

Wait: It says "12 blue, 7 red, 5 white, and 4 black" → That’s 12+7+5+4 = 28 marbles, but it says "a bag with 10 marbles". So likely a typo.

But the problem states: *"Thomas had a bag with 10 marbles: 12 blue, 7 red..."* — that can't be.

Possibility: Maybe it was meant to be 1 blue, 2 red, 5 white, and 2 black? But let's re-read carefully.

Actually, it says:
> "Thomas had a bag with 10 marbles: 12 blue, 7 red, 5 white, and 4 black."

This is impossible because 12+7+5+4 = 28 ≠ 10.

So likely a typo. Perhaps the numbers are wrong.

But wait — maybe it's not 10 marbles, or the numbers are incorrect.

Alternatively, perhaps the bag has 10 marbles, and the counts are different.

But as written, this is inconsistent.

Let me assume it's a typo and suppose the correct total is 28 marbles: 12 blue, 7 red, 5 white, 4 black.

Then proceed accordingly.

But the problem says “a bag with 10 marbles” — so maybe it's a mistake in the number of marbles.

Let’s look at part (a): “He took a marble. What is the probability that it is blue?”

If there are 12 blue marbles, but only 10 total, impossible.

So likely, the numbers are wrong.

Another possibility: Maybe the bag has 28 marbles, and the sentence should say “28 marbles”.

We’ll assume that the total is 28 marbles, and the counts are:
- Blue: 12
- Red: 7
- White: 5
- Black: 4

Total = 12+7+5+4 = 28 ✔

Now solve:

(a) Probability of drawing a blue marble:
$$
P(\text{blue}) = \frac{12}{28} = \frac{3}{7}
$$

(b) Probability of drawing a non-blue marble:
That is, red, white, or black.
$$
P(\text{not blue}) = 1 - P(\text{blue}) = 1 - \frac{12}{28} = \frac{16}{28} = \frac{4}{7}
$$

Or directly: non-blue = 7+5+4 = 16 → $ \frac{16}{28} = \frac{4}{7} $

✔ Answer:
- (a) $ \frac{3}{7} $
- (b) $ \frac{4}{7} $

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Problem 3:

> Julie was given two bags full of gumballs. There were 10 chocolate bars and 5 mint candies in Bag A. There were 7 chocolate bars and 3 strawberry candies in Bag B. She wanted to eat a chocolate bar. From which bag should she take the candy so that the probability of eating a chocolate bar was greatest?

Let’s compute the probability of picking a chocolate bar from each bag.

Bag A:
- Chocolate: 10
- Mint: 5
- Total: 15
$$
P(\text{chocolate | Bag A}) = \frac{10}{15} = \frac{2}{3} \approx 0.6667
$$

Bag B:
- Chocolate: 7
- Strawberry: 3
- Total: 10
$$
P(\text{chocolate | Bag B}) = \frac{7}{10} = 0.7
$$

Compare:
- Bag A: $ \frac{2}{3} \approx 0.6667 $
- Bag B: $ 0.7 $

So Bag B has higher probability.

✔ Answer: She should take from Bag B.

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Problem 4:

> Mike’s class made a project about pets. There were 21 students in his class. Nine had dogs, five had only cats, three had cats and hamsters, and the rest had no pets at all. They randomly picked one student.

Let’s break down the data:

- Total students: 21
- Dogs: 9
- Only cats: 5
- Cats and hamsters: 3
- Rest: no pets

Note: “Three had cats and hamsters” — does this mean they have both cats and hamsters? Yes.

But do these people also have cats? Yes, so we need to be careful.

Let’s count how many students have pets:

- Dogs only: 9
- Only cats: 5
- Cats and hamsters: 3

Are these groups disjoint?

Yes — "only cats" means no other pets; "cats and hamsters" means both, so not "only cats".

So total with pets:
- Dogs: 9
- Only cats: 5
- Cats and hamsters: 3

Total pet owners = 9 + 5 + 3 = 17

No pets = 21 - 17 = 4

Now:

(a) What was the probability that the student had a dog?
$$
P(\text{dog}) = \frac{9}{21} = \frac{3}{7}
$$

(b) What was the probability that the student had at least one pet?
At least one pet = 17 students
$$
P(\text{at least one pet}) = \frac{17}{21}
$$

✔ Answer:
- (a) $ \frac{3}{7} $
- (b) $ \frac{17}{21} $

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Problem 5:

> Mr. Johnson taught a math class for 25 students under the age of 18. He randomly chose one of them. What was the probability that the student was under twelve?

Wait: The class has 25 students under the age of 18, and he chooses one at random.

But we are told nothing about how many are under 12.

The question asks for the probability that the student is under 12, but we don’t have data on how many are under 12.

So unless we assume something, we cannot answer.

But the problem says: “Mr. Johnson taught a math class for 25 students under the age of 18.”

So all 25 are under 18.

But we don’t know how many are under 12.

So insufficient information.

But perhaps it’s implied that “under 12” is a subset, but no numbers are given.

Wait — maybe the problem is missing data.

But let’s read again: “he randomly chose one of them. What was the probability that the student was under twelve?”

But no breakdown of ages.

So unless we assume something, we can’t solve.

But perhaps it’s a trick — maybe the class is all under 18, but we don’t know how many under 12.

So we cannot determine the probability without more info.

But maybe the problem is misstated.

Alternatively, perhaps “under the age of 18” is just describing the class, and we’re supposed to assume something else?

But still, we need numbers.

Wait — could it be that the question is asking for the probability that a student is under 12, given that all are under 18? But without knowing how many are under 12, we can’t compute it.

So this problem is unsolvable as stated.

But maybe there's a typo — perhaps it said “12 students under 12”, but it doesn’t.

So unless additional data is provided, we must say:

✘ Insufficient information.

But let’s see if we missed something.

Wait — maybe the phrase “under the age of 18” is just descriptive, and the class has 25 students, all under 18, but we don’t know how many are under 12.

So yes, no way to compute.

Unless the problem meant to say something like “12 of them are under 12”, but it doesn’t.

So cannot solve.

But perhaps the answer is expected to be “unknown” or “not enough information”.

✔ Answer: Not enough information to determine the probability.

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Problem 6:

> The computer disk listed track number 12 songs. The first four were rock music. Tracks numbered 5 through 12 were ballads. She selected the random function in her Compact Disk Player. When is the probability of first listening to a ballad?

Total tracks: 12

- Rock: tracks 1–4 → 4 songs
- Ballads: tracks 5–12 → 8 songs

She selects a song at random.

So probability of selecting a ballad:
$$
P(\text{ballad}) = \frac{8}{12} = \frac{2}{3}
$$

✔ Answer: $ \frac{2}{3} $

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✔ Final Answers Summary:

1. (a) $ \frac{21}{32} $, (b) $ \frac{11}{32} $
2. (a) $ \frac{3}{7} $, (b) $ \frac{4}{7} $
3. Bag B
4. (a) $ \frac{3}{7} $, (b) $ \frac{17}{21} $
5. Not enough information
6. $ \frac{2}{3} $

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⚠️ Note: Problem 2 has an inconsistency (12+7+5+4=28 ≠ 10). We assumed the total was 28. If the total is truly 10, then the numbers must be wrong. Please double-check the original image.

Similarly, Problem 5 lacks sufficient data.

Let me know if you'd like me to adjust based on corrected values.