Mutually exclusive events - Free Printable
Educational worksheet: Mutually exclusive events. Download and print for classroom or home learning activities.
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Step-by-step solution for: Mutually exclusive events
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Show Answer Key & Explanations
Step-by-step solution for: Mutually exclusive events
Here are the step-by-step solutions for each problem on your worksheet.
ME or non: ME (Mutually Exclusive)
*Reasoning:* A single coin cannot be both a penny and a dime at the same time.
Probability Calculation:
1. Find the total number of coins (Sample Space):
$4 \text{ pennies} + 3 \text{ nickels} + 6 \text{ dimes} = 13 \text{ total coins}$.
2. Find the number of favorable outcomes:
We want a penny OR a dime.
$4 \text{ pennies} + 6 \text{ dimes} = 10 \text{ favorable coins}$.
3. Calculate Probability:
$$P(\text{penny or dime}) = \frac{10}{13}$$
---
Total people in the table: 27
a. Probability of calling on a blond-haired person or a boy
ME or non: non (Non-Mutually Exclusive)
*Reasoning:* It is possible to be both a boy AND have blond hair (there are 6 boys with blond hair). Because they overlap, they are not mutually exclusive.
Probability Calculation:
* Total Blond-Haired people: 11
* Total Boys: 15
* People who are BOTH (Blond Boys): 6
* Formula: $P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)$
$$\frac{11}{27} + \frac{15}{27} - \frac{6}{27} = \frac{20}{27}$$
b. Probability of calling on a blond-haired girl or a boy
ME or non: ME (Mutually Exclusive)
*Reasoning:* A person cannot be a "girl" and a "boy" at the same time. Even though some girls are blond, the category "Boy" does not overlap with "Girl."
Probability Calculation:
* Blond-Haired Girls: 5
* Total Boys: 15
* Since they don't overlap, just add them up:
$5 + 15 = 20$
$$P(\text{blond girl or boy}) = \frac{20}{27}$$
---
ME or non: non (Non-Mutually Exclusive)
*Reasoning:* There is a Queen of Diamonds. It is possible to draw a card that is both a diamond and a queen.
Probability Calculation:
* Total cards: 52
* Total Diamonds: 13
* Total Queens: 4
* Queen of Diamonds (overlap): 1
* Formula: $\frac{13}{52} + \frac{4}{52} - \frac{1}{52} = \frac{16}{52}$
* Simplify the fraction (divide top and bottom by 4):
$$\frac{4}{13}$$
---
ME or non: non (Non-Mutually Exclusive)
*Reasoning:* You could roll a 4 on the first die, the second die, or both. The event "First die is 4" and "Second die is 4" can happen together (rolling double 4s).
Probability Calculation:
* Total outcomes for two dice: $6 \times 6 = 36$.
* Outcomes with a 4 on the first die: $(4,1), (4,2), (4,3), (4,4), (4,5), (4,6)$ → 6 outcomes.
* Outcomes with a 4 on the second die: $(1,4), (2,4), (3,4), (4,4), (5,4), (6,4)$ → 6 outcomes.
* Note that $(4,4)$ is counted in both lists.
* Unique favorable outcomes: $6 + 6 - 1 = 11$.
$$P(\text{at least one 4}) = \frac{11}{36}$$
---
ME or non: non (Non-Mutually Exclusive)
*Reasoning:* There are red Aces (Ace of Hearts and Ace of Diamonds).
Probability Calculation:
* Total cards: 52
* Total Aces: 4
* Total Red cards: 26
* Red Aces (overlap): 2
* Formula: $\frac{4}{52} + \frac{26}{52} - \frac{2}{52} = \frac{28}{52}$
* Simplify (divide by 4):
$$\frac{7}{13}$$
---
ME or non: ME (Mutually Exclusive)
*Reasoning:* The sum of the dice can only be one number at a time. It cannot be 6 and 9 simultaneously.
Probability Calculation:
* Total outcomes: 36
* Ways to get a sum of 6: $(1,5), (2,4), (3,3), (4,2), (5,1)$ → 5 ways.
* Ways to get a sum of 9: $(3,6), (4,5), (5,4), (6,3)$ → 4 ways.
* Total favorable outcomes: $5 + 4 = 9$.
$$P(\text{sum 6 or 9}) = \frac{9}{36} = \frac{1}{4}$$
---
Total cards: 14
a) Selecting a prime number or a multiple of four
ME or non: ME (Mutually Exclusive)
*Reasoning:* Let's list them.
Primes (1-14): 2, 3, 5, 7, 11, 13.
Multiples of 4 (1-14): 4, 8, 12.
There are no numbers that appear on both lists.
Probability:
* Count: 6 primes + 3 multiples of 4 = 9 favorable cards.
$$P = \frac{9}{14}$$
b) Selecting a multiple of two or a multiple of three
ME or non: non (Non-Mutually Exclusive)
*Reasoning:* Some numbers are multiples of both 2 and 3 (like 6 and 12).
Probability:
* Multiples of 2: 2, 4, 6, 8, 10, 12, 14 (7 numbers).
* Multiples of 3: 3, 6, 9, 12 (4 numbers).
* Overlap (multiples of 6): 6, 12 (2 numbers).
* Calculation: $7 + 4 - 2 = 9$ favorable cards.
$$P = \frac{9}{14}$$
c) Selecting a 3 or a 4
ME or non: ME (Mutually Exclusive)
*Reasoning:* A single card cannot be the number 3 and the number 4 at the same time.
Probability:
* Favorable cards: 2 (the card '3' and the card '4').
$$P = \frac{2}{14} = \frac{1}{7}$$
d) Selecting an 8 or a number less than 8
ME or non: ME (Mutually Exclusive)
*Reasoning:* The number 8 is not "less than 8". They are separate groups.
Probability:
* Card '8': 1 card.
* Numbers less than 8: 1, 2, 3, 4, 5, 6, 7 (7 cards).
* Total favorable: $1 + 7 = 8$.
$$P = \frac{8}{14} = \frac{4}{7}$$
──────────────────────────────────────
Final Answer:
1. ME; $\frac{10}{13}$
2a. non; $\frac{20}{27}$
2b. ME; $\frac{20}{27}$
3. non; $\frac{4}{13}$
4. non; $\frac{11}{36}$
5. non; $\frac{7}{13}$
6. ME; $\frac{1}{4}$
7a. ME; $\frac{9}{14}$
7b. non; $\frac{9}{14}$
7c. ME; $\frac{1}{7}$
7d. ME; $\frac{4}{7}$
1. Choosing a penny or a dime
ME or non: ME (Mutually Exclusive)
*Reasoning:* A single coin cannot be both a penny and a dime at the same time.
Probability Calculation:
1. Find the total number of coins (Sample Space):
$4 \text{ pennies} + 3 \text{ nickels} + 6 \text{ dimes} = 13 \text{ total coins}$.
2. Find the number of favorable outcomes:
We want a penny OR a dime.
$4 \text{ pennies} + 6 \text{ dimes} = 10 \text{ favorable coins}$.
3. Calculate Probability:
$$P(\text{penny or dime}) = \frac{10}{13}$$
---
2. Using the Table
Total people in the table: 27
a. Probability of calling on a blond-haired person or a boy
ME or non: non (Non-Mutually Exclusive)
*Reasoning:* It is possible to be both a boy AND have blond hair (there are 6 boys with blond hair). Because they overlap, they are not mutually exclusive.
Probability Calculation:
* Total Blond-Haired people: 11
* Total Boys: 15
* People who are BOTH (Blond Boys): 6
* Formula: $P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)$
$$\frac{11}{27} + \frac{15}{27} - \frac{6}{27} = \frac{20}{27}$$
b. Probability of calling on a blond-haired girl or a boy
ME or non: ME (Mutually Exclusive)
*Reasoning:* A person cannot be a "girl" and a "boy" at the same time. Even though some girls are blond, the category "Boy" does not overlap with "Girl."
Probability Calculation:
* Blond-Haired Girls: 5
* Total Boys: 15
* Since they don't overlap, just add them up:
$5 + 15 = 20$
$$P(\text{blond girl or boy}) = \frac{20}{27}$$
---
3. Drawing a diamond or queen from a deck
ME or non: non (Non-Mutually Exclusive)
*Reasoning:* There is a Queen of Diamonds. It is possible to draw a card that is both a diamond and a queen.
Probability Calculation:
* Total cards: 52
* Total Diamonds: 13
* Total Queens: 4
* Queen of Diamonds (overlap): 1
* Formula: $\frac{13}{52} + \frac{4}{52} - \frac{1}{52} = \frac{16}{52}$
* Simplify the fraction (divide top and bottom by 4):
$$\frac{4}{13}$$
---
4. Tossing two dice showing at least one 4
ME or non: non (Non-Mutually Exclusive)
*Reasoning:* You could roll a 4 on the first die, the second die, or both. The event "First die is 4" and "Second die is 4" can happen together (rolling double 4s).
Probability Calculation:
* Total outcomes for two dice: $6 \times 6 = 36$.
* Outcomes with a 4 on the first die: $(4,1), (4,2), (4,3), (4,4), (4,5), (4,6)$ → 6 outcomes.
* Outcomes with a 4 on the second die: $(1,4), (2,4), (3,4), (4,4), (5,4), (6,4)$ → 6 outcomes.
* Note that $(4,4)$ is counted in both lists.
* Unique favorable outcomes: $6 + 6 - 1 = 11$.
$$P(\text{at least one 4}) = \frac{11}{36}$$
---
5. Selecting an ace or a red card
ME or non: non (Non-Mutually Exclusive)
*Reasoning:* There are red Aces (Ace of Hearts and Ace of Diamonds).
Probability Calculation:
* Total cards: 52
* Total Aces: 4
* Total Red cards: 26
* Red Aces (overlap): 2
* Formula: $\frac{4}{52} + \frac{26}{52} - \frac{2}{52} = \frac{28}{52}$
* Simplify (divide by 4):
$$\frac{7}{13}$$
---
6. Two dice showing a sum of 6 or a sum of 9
ME or non: ME (Mutually Exclusive)
*Reasoning:* The sum of the dice can only be one number at a time. It cannot be 6 and 9 simultaneously.
Probability Calculation:
* Total outcomes: 36
* Ways to get a sum of 6: $(1,5), (2,4), (3,3), (4,2), (5,1)$ → 5 ways.
* Ways to get a sum of 9: $(3,6), (4,5), (5,4), (6,3)$ → 4 ways.
* Total favorable outcomes: $5 + 4 = 9$.
$$P(\text{sum 6 or 9}) = \frac{9}{36} = \frac{1}{4}$$
---
7. Bag with cards numbered 1 to 14
Total cards: 14
a) Selecting a prime number or a multiple of four
ME or non: ME (Mutually Exclusive)
*Reasoning:* Let's list them.
Primes (1-14): 2, 3, 5, 7, 11, 13.
Multiples of 4 (1-14): 4, 8, 12.
There are no numbers that appear on both lists.
Probability:
* Count: 6 primes + 3 multiples of 4 = 9 favorable cards.
$$P = \frac{9}{14}$$
b) Selecting a multiple of two or a multiple of three
ME or non: non (Non-Mutually Exclusive)
*Reasoning:* Some numbers are multiples of both 2 and 3 (like 6 and 12).
Probability:
* Multiples of 2: 2, 4, 6, 8, 10, 12, 14 (7 numbers).
* Multiples of 3: 3, 6, 9, 12 (4 numbers).
* Overlap (multiples of 6): 6, 12 (2 numbers).
* Calculation: $7 + 4 - 2 = 9$ favorable cards.
$$P = \frac{9}{14}$$
c) Selecting a 3 or a 4
ME or non: ME (Mutually Exclusive)
*Reasoning:* A single card cannot be the number 3 and the number 4 at the same time.
Probability:
* Favorable cards: 2 (the card '3' and the card '4').
$$P = \frac{2}{14} = \frac{1}{7}$$
d) Selecting an 8 or a number less than 8
ME or non: ME (Mutually Exclusive)
*Reasoning:* The number 8 is not "less than 8". They are separate groups.
Probability:
* Card '8': 1 card.
* Numbers less than 8: 1, 2, 3, 4, 5, 6, 7 (7 cards).
* Total favorable: $1 + 7 = 8$.
$$P = \frac{8}{14} = \frac{4}{7}$$
──────────────────────────────────────
Final Answer:
1. ME; $\frac{10}{13}$
2a. non; $\frac{20}{27}$
2b. ME; $\frac{20}{27}$
3. non; $\frac{4}{13}$
4. non; $\frac{11}{36}$
5. non; $\frac{7}{13}$
6. ME; $\frac{1}{4}$
7a. ME; $\frac{9}{14}$
7b. non; $\frac{9}{14}$
7c. ME; $\frac{1}{7}$
7d. ME; $\frac{4}{7}$
Parent Tip: Review the logic above to help your child master the concept of probability mutually exclusive events worksheet answers.
