Independent Events Worksheet featuring ten math problems on probability.
Independent Events Worksheet with ten probability problems involving dice, cards, and random selection.
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Step-by-step solution for: Fillable Online Independent Events Five Worksheet Pack Fax Email ...
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Step-by-step solution for: Fillable Online Independent Events Five Worksheet Pack Fax Email ...
Here are the step-by-step solutions for each problem on the worksheet.
1. Probability of a number from 20 to 30 being divisible by 2 and then 15 (with replacement)
* Step 1: List the numbers from 20 to 30: $\{20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30\}$. There are 11 numbers in total.
* Step 2: Find numbers divisible by 2 (even numbers): $\{20, 22, 24, 26, 28, 30\}$. There are 6 even numbers.
* Probability (divisible by 2) = $6/11$.
* Step 3: Find numbers divisible by 15: Only $\{30\}$ is divisible by 15 in this range. There is 1 such number.
* Probability (divisible by 15) = $1/11$.
* Step 4: Since the selection is "with replacement," we multiply the probabilities.
* Calculation: $\frac{6}{11} \times \frac{1}{11} = \frac{6}{121}$.
2. Probability of getting a Club from a deck
* Step 1: A standard deck has 52 cards.
* Step 2: There are 4 suits (Hearts, Diamonds, Clubs, Spades), and each has 13 cards. So, there are 13 Clubs.
* Step 3: Divide the number of Clubs by the total cards.
* Calculation: $\frac{13}{52}$ simplifies to $\frac{1}{4}$.
3. Probability of picking a yellow ball (without replacement)
* Step 1: Count the total balls: $20 \text{ blue} + 10 \text{ yellow} + 8 \text{ red} + 4 \text{ white} = 42$ balls.
* Step 2: Identify the target: Yellow balls = 10.
* Step 3: Create the fraction: $\frac{10}{42}$.
* Step 4: Simplify the fraction by dividing top and bottom by 2.
* Calculation: $\frac{5}{21}$.
4. Probability of rolling an odd number OR a six
* Step 1: A die has sides $\{1, 2, 3, 4, 5, 6\}$. Total outcomes = 6.
* Step 2: Odd numbers are $\{1, 3, 5\}$. That is 3 outcomes.
* Step 3: The number 6 is one outcome.
* Step 4: Check for overlap: 6 is not odd, so these events don't overlap. We just add them up.
* Favorable outcomes: $\{1, 3, 5, 6\}$. That is 4 outcomes.
* Calculation: $\frac{4}{6}$ simplifies to $\frac{2}{3}$.
5. Probability of King then Black Jack (without replacement)
* Step 1: Probability of first card being a King. There are 4 Kings in 52 cards.
* $P(\text{King}) = \frac{4}{52} = \frac{1}{13}$.
* Step 2: Probability of second card being a Black Jack. Since we didn't replace the first card, there are 51 cards left. There are 2 Black Jacks (Jack of Spades, Jack of Clubs).
* $P(\text{Black Jack}) = \frac{2}{51}$.
* Step 3: Multiply the probabilities.
* Calculation: $\frac{1}{13} \times \frac{2}{51} = \frac{2}{663}$.
6. Probability integer 1 through 15 is even
* Step 1: Total integers = 15.
* Step 2: Even integers are $\{2, 4, 6, 8, 10, 12, 14\}$. Count = 7.
* Step 3: Create the fraction.
* Calculation: $\frac{7}{15}$.
7. Probability of Blue then Red (without replacement)
* Step 1: Total marbles: $20 \text{ blue} + 15 \text{ red} = 35$ marbles.
* Step 2: Pick Blue first. There are 20 blue marbles.
* $P(\text{Blue}) = \frac{20}{35}$.
* Step 3: Pick Red second. One marble is gone, so 34 remain. There are still 15 red marbles.
* $P(\text{Red}) = \frac{15}{34}$.
* Step 4: Multiply.
* $\frac{20}{35} \times \frac{15}{34}$.
* Simplify before multiplying: $\frac{20}{35}$ becomes $\frac{4}{7}$. $\frac{15}{34}$ stays same.
* $\frac{4}{7} \times \frac{15}{34} = \frac{60}{238}$.
* Simplify $\frac{60}{238}$ by dividing by 2: $\frac{30}{119}$.
8. Probability of English paper then History paper (without replacement)
* Step 1: Total papers: $4 \text{ science} + 6 \text{ English} + 3 \text{ history} = 13$ papers.
* Step 2: Pick English first. There are 6 English papers.
* $P(\text{English}) = \frac{6}{13}$.
* Step 3: Pick History second. One paper is gone, so 12 remain. There are 3 history papers.
* $P(\text{History}) = \frac{3}{12} = \frac{1}{4}$.
* Step 4: Multiply.
* Calculation: $\frac{6}{13} \times \frac{1}{4} = \frac{6}{52}$.
* Simplify: $\frac{3}{26}$.
9. Probability of Brown card then Red card (with replacement)
* Step 1: Total cards: $12 \text{ brown} + 14 \text{ red} + 12 \text{ green} = 38$ cards.
* Step 2: Pick Brown first.
* $P(\text{Brown}) = \frac{12}{38} = \frac{6}{19}$.
* Step 3: Pick Red second. Because it is with replacement, the total goes back to 38.
* $P(\text{Red}) = \frac{14}{38} = \frac{7}{19}$.
* Step 4: Multiply.
* Calculation: $\frac{6}{19} \times \frac{7}{19} = \frac{42}{361}$.
10. Probability of choosing three girls sequentially
* Step 1: Total students: $4 \text{ boys} + 6 \text{ girls} = 10$ students.
* Step 2: Choose 1st girl. There are 6 girls out of 10.
* $P(1) = \frac{6}{10} = \frac{3}{5}$.
* Step 3: Choose 2nd girl. One girl is picked, so 9 students remain, and 5 girls remain.
* $P(2) = \frac{5}{9}$.
* Step 4: Choose 3rd girl. Two girls are picked, so 8 students remain, and 4 girls remain.
* $P(3) = \frac{4}{8} = \frac{1}{2}$.
* Step 5: Multiply all three probabilities.
* Calculation: $\frac{3}{5} \times \frac{5}{9} \times \frac{1}{2}$.
* Cancel the 5s: $\frac{3}{1} \times \frac{1}{9} \times \frac{1}{2} = \frac{3}{18}$.
* Simplify: $\frac{1}{6}$.
Final Answer:
1. 6/121
2. 1/4
3. 5/21
4. 2/3
5. 2/663
6. 7/15
7. 30/119
8. 3/26
9. 42/361
10. 1/6
1. Probability of a number from 20 to 30 being divisible by 2 and then 15 (with replacement)
* Step 1: List the numbers from 20 to 30: $\{20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30\}$. There are 11 numbers in total.
* Step 2: Find numbers divisible by 2 (even numbers): $\{20, 22, 24, 26, 28, 30\}$. There are 6 even numbers.
* Probability (divisible by 2) = $6/11$.
* Step 3: Find numbers divisible by 15: Only $\{30\}$ is divisible by 15 in this range. There is 1 such number.
* Probability (divisible by 15) = $1/11$.
* Step 4: Since the selection is "with replacement," we multiply the probabilities.
* Calculation: $\frac{6}{11} \times \frac{1}{11} = \frac{6}{121}$.
2. Probability of getting a Club from a deck
* Step 1: A standard deck has 52 cards.
* Step 2: There are 4 suits (Hearts, Diamonds, Clubs, Spades), and each has 13 cards. So, there are 13 Clubs.
* Step 3: Divide the number of Clubs by the total cards.
* Calculation: $\frac{13}{52}$ simplifies to $\frac{1}{4}$.
3. Probability of picking a yellow ball (without replacement)
* Step 1: Count the total balls: $20 \text{ blue} + 10 \text{ yellow} + 8 \text{ red} + 4 \text{ white} = 42$ balls.
* Step 2: Identify the target: Yellow balls = 10.
* Step 3: Create the fraction: $\frac{10}{42}$.
* Step 4: Simplify the fraction by dividing top and bottom by 2.
* Calculation: $\frac{5}{21}$.
4. Probability of rolling an odd number OR a six
* Step 1: A die has sides $\{1, 2, 3, 4, 5, 6\}$. Total outcomes = 6.
* Step 2: Odd numbers are $\{1, 3, 5\}$. That is 3 outcomes.
* Step 3: The number 6 is one outcome.
* Step 4: Check for overlap: 6 is not odd, so these events don't overlap. We just add them up.
* Favorable outcomes: $\{1, 3, 5, 6\}$. That is 4 outcomes.
* Calculation: $\frac{4}{6}$ simplifies to $\frac{2}{3}$.
5. Probability of King then Black Jack (without replacement)
* Step 1: Probability of first card being a King. There are 4 Kings in 52 cards.
* $P(\text{King}) = \frac{4}{52} = \frac{1}{13}$.
* Step 2: Probability of second card being a Black Jack. Since we didn't replace the first card, there are 51 cards left. There are 2 Black Jacks (Jack of Spades, Jack of Clubs).
* $P(\text{Black Jack}) = \frac{2}{51}$.
* Step 3: Multiply the probabilities.
* Calculation: $\frac{1}{13} \times \frac{2}{51} = \frac{2}{663}$.
6. Probability integer 1 through 15 is even
* Step 1: Total integers = 15.
* Step 2: Even integers are $\{2, 4, 6, 8, 10, 12, 14\}$. Count = 7.
* Step 3: Create the fraction.
* Calculation: $\frac{7}{15}$.
7. Probability of Blue then Red (without replacement)
* Step 1: Total marbles: $20 \text{ blue} + 15 \text{ red} = 35$ marbles.
* Step 2: Pick Blue first. There are 20 blue marbles.
* $P(\text{Blue}) = \frac{20}{35}$.
* Step 3: Pick Red second. One marble is gone, so 34 remain. There are still 15 red marbles.
* $P(\text{Red}) = \frac{15}{34}$.
* Step 4: Multiply.
* $\frac{20}{35} \times \frac{15}{34}$.
* Simplify before multiplying: $\frac{20}{35}$ becomes $\frac{4}{7}$. $\frac{15}{34}$ stays same.
* $\frac{4}{7} \times \frac{15}{34} = \frac{60}{238}$.
* Simplify $\frac{60}{238}$ by dividing by 2: $\frac{30}{119}$.
8. Probability of English paper then History paper (without replacement)
* Step 1: Total papers: $4 \text{ science} + 6 \text{ English} + 3 \text{ history} = 13$ papers.
* Step 2: Pick English first. There are 6 English papers.
* $P(\text{English}) = \frac{6}{13}$.
* Step 3: Pick History second. One paper is gone, so 12 remain. There are 3 history papers.
* $P(\text{History}) = \frac{3}{12} = \frac{1}{4}$.
* Step 4: Multiply.
* Calculation: $\frac{6}{13} \times \frac{1}{4} = \frac{6}{52}$.
* Simplify: $\frac{3}{26}$.
9. Probability of Brown card then Red card (with replacement)
* Step 1: Total cards: $12 \text{ brown} + 14 \text{ red} + 12 \text{ green} = 38$ cards.
* Step 2: Pick Brown first.
* $P(\text{Brown}) = \frac{12}{38} = \frac{6}{19}$.
* Step 3: Pick Red second. Because it is with replacement, the total goes back to 38.
* $P(\text{Red}) = \frac{14}{38} = \frac{7}{19}$.
* Step 4: Multiply.
* Calculation: $\frac{6}{19} \times \frac{7}{19} = \frac{42}{361}$.
10. Probability of choosing three girls sequentially
* Step 1: Total students: $4 \text{ boys} + 6 \text{ girls} = 10$ students.
* Step 2: Choose 1st girl. There are 6 girls out of 10.
* $P(1) = \frac{6}{10} = \frac{3}{5}$.
* Step 3: Choose 2nd girl. One girl is picked, so 9 students remain, and 5 girls remain.
* $P(2) = \frac{5}{9}$.
* Step 4: Choose 3rd girl. Two girls are picked, so 8 students remain, and 4 girls remain.
* $P(3) = \frac{4}{8} = \frac{1}{2}$.
* Step 5: Multiply all three probabilities.
* Calculation: $\frac{3}{5} \times \frac{5}{9} \times \frac{1}{2}$.
* Cancel the 5s: $\frac{3}{1} \times \frac{1}{9} \times \frac{1}{2} = \frac{3}{18}$.
* Simplify: $\frac{1}{6}$.
Final Answer:
1. 6/121
2. 1/4
3. 5/21
4. 2/3
5. 2/663
6. 7/15
7. 30/119
8. 3/26
9. 42/361
10. 1/6
Parent Tip: Review the logic above to help your child master the concept of independent events worksheet.
