probability rules practice worksheet - fall 2025 - statistics - room 404\nfor each problem listed below…

probability rules practice worksheet - fall 2025 - statistics - room 404\nfor each problem listed below, find p(a), p(b), p(a), p(b), p(a∩b), p(a∪b), and m.e.?\na number is randomly chosen from 1 to 20.\n1. a: # greater than 15 b: multiple of 4\n2. a: even # b: 11\na single card is randomly chosen from a deck.\n3. a: black b: jack\n4. a: face b: club\n5. a: ace b: number\ntwo dice are rolled.\n6. a: sum of 8 b: at least one of the dice is a 5\n7. a: sum of 5 or lower is rolled b: the first die rolled is a 6\n8. given the following venn diagram and associated probabilities of sample points. p(1)=.28 p(2)=.12 p(3)=.18 p(4)=.42\n9. given the following chart(a: female, b: vote no)\nfemale male\nyes 24 9 33\nno 9 18 27\n33 27 60

probability rules practice worksheet - fall 2025 - statistics - room 404\nfor each problem listed below, find p(a), p(b), p(a), p(b), p(a∩b), p(a∪b), and m.e.?\na number is randomly chosen from 1 to 20.\n1. a: # greater than 15 b: multiple of 4\n2. a: even # b: 11\na single card is randomly chosen from a deck.\n3. a: black b: jack\n4. a: face b: club\n5. a: ace b: number\ntwo dice are rolled.\n6. a: sum of 8 b: at least one of the dice is a 5\n7. a: sum of 5 or lower is rolled b: the first die rolled is a 6\n8. given the following venn diagram and associated probabilities of sample points. p(1)=.28 p(2)=.12 p(3)=.18 p(4)=.42\n9. given the following chart(a: female, b: vote no)\nfemale male\nyes 24 9 33\nno 9 18 27\n33 27 60

Answer

Explanation:

Step1: Calculate P(A)

There are 5 numbers greater than 15 in the set {1, 2, ..., 20}, so $P(A)=\frac{5}{20}=\frac{1}{4}$.

Step2: Calculate P(B)

There are 5 multiples of 4 in the set {1, 2, ..., 20} (4, 8, 12, 16, 20), so $P(B)=\frac{5}{20}=\frac{1}{4}$.

Step3: Calculate P(A')

$P(A') = 1 - P(A)=1-\frac{1}{4}=\frac{3}{4}$.

Step4: Calculate P(B')

$P(B') = 1 - P(B)=1 - \frac{1}{4}=\frac{3}{4}$.

Step5: Calculate P(A∩B)

The numbers that are greater than 15 and multiples of 4 are 16 and 20, so $P(A\cap B)=\frac{2}{20}=\frac{1}{10}$.

Step6: Calculate P(A∪B)

Using the formula $P(A\cup B)=P(A)+P(B)-P(A\cap B)$, we have $P(A\cup B)=\frac{1}{4}+\frac{1}{4}-\frac{1}{10}=\frac{5 + 5- 2}{20}=\frac{8}{20}=\frac{2}{5}$.

Answer:

$P(A)=\frac{1}{4}$, $P(B)=\frac{1}{4}$, $P(A')=\frac{3}{4}$, $P(B')=\frac{3}{4}$, $P(A\cap B)=\frac{1}{10}$, $P(A\cup B)=\frac{2}{5}$

Explanation:

Step1: Calculate P(A)

There are 10 even - numbered elements in the set {1, 2, ..., 20}, so $P(A)=\frac{10}{20}=\frac{1}{2}$.

Step2: Calculate P(B)

There is 1 element equal to 11 in the set {1, 2, ..., 20}, so $P(B)=\frac{1}{20}$.

Step3: Calculate P(A')

$P(A') = 1 - P(A)=1-\frac{1}{2}=\frac{1}{2}$.

Step4: Calculate P(B')

$P(B') = 1 - P(B)=1-\frac{1}{20}=\frac{19}{20}$.

Step5: Calculate P(A∩B)

Since 11 is odd, $P(A\cap B) = 0$.

Step6: Calculate P(A∪B)

Using the formula $P(A\cup B)=P(A)+P(B)-P(A\cap B)$, we have $P(A\cup B)=\frac{1}{2}+\frac{1}{20}-0=\frac{10 + 1}{20}=\frac{11}{20}$.

Answer:

$P(A)=\frac{1}{2}$, $P(B)=\frac{1}{20}$, $P(A')=\frac{1}{2}$, $P(B')=\frac{19}{20}$, $P(A\cap B)=0$, $P(A\cup B)=\frac{11}{20}$

Explanation:

Step1: Calculate P(A)

There are 26 black cards in a standard deck of 52 cards, so $P(A)=\frac{26}{52}=\frac{1}{2}$.

Step2: Calculate P(B)

There are 4 Jacks in a standard deck of 52 cards, so $P(B)=\frac{4}{52}=\frac{1}{13}$.

Step3: Calculate P(A')

$P(A') = 1 - P(A)=1-\frac{1}{2}=\frac{1}{2}$.

Step4: Calculate P(B')

$P(B') = 1 - P(B)=1-\frac{1}{13}=\frac{12}{13}$.

Step5: Calculate P(A∩B)

There are 2 black Jacks (Jack of spades and Jack of clubs), so $P(A\cap B)=\frac{2}{52}=\frac{1}{26}$.

Step6: Calculate P(A∪B)

Using the formula $P(A\cup B)=P(A)+P(B)-P(A\cap B)$, we have $P(A\cup B)=\frac{1}{2}+\frac{1}{13}-\frac{1}{26}=\frac{13 + 2-1}{26}=\frac{7}{13}$.

Answer:

$P(A)=\frac{1}{2}$, $P(B)=\frac{1}{13}$, $P(A')=\frac{1}{2}$, $P(B')=\frac{12}{13}$, $P(A\cap B)=\frac{1}{26}$, $P(A\cup B)=\frac{7}{13}$

Explanation:

Step1: Calculate P(A)

There are 12 face - cards in a standard deck of 52 cards, so $P(A)=\frac{12}{52}=\frac{3}{13}$.

Step2: Calculate P(B)

There are 13 clubs in a standard deck of 52 cards, so $P(B)=\frac{13}{52}=\frac{1}{4}$.

Step3: Calculate P(A')

$P(A') = 1 - P(A)=1-\frac{3}{13}=\frac{10}{13}$.

Step4: Calculate P(B')

$P(B') = 1 - P(B)=1-\frac{1}{4}=\frac{3}{4}$.

Step5: Calculate P(A∩B)

There are 3 face - cards that are clubs (Jack, Queen, King of clubs), so $P(A\cap B)=\frac{3}{52}$.

Step6: Calculate P(A∪B)

Using the formula $P(A\cup B)=P(A)+P(B)-P(A\cap B)$, we have $P(A\cup B)=\frac{3}{13}+\frac{1}{4}-\frac{3}{52}=\frac{12 + 13-3}{52}=\frac{11}{26}$.

Answer:

$P(A)=\frac{3}{13}$, $P(B)=\frac{1}{4}$, $P(A')=\frac{10}{13}$, $P(B')=\frac{3}{4}$, $P(A\cap B)=\frac{3}{52}$, $P(A\cup B)=\frac{11}{26}$

Explanation:

Step1: Calculate P(A)

There are 4 Aces in a standard deck of 52 cards, so $P(A)=\frac{4}{52}=\frac{1}{13}$.

Step2: Calculate P(B)

There are 36 numbered cards (2 - 10 in each suit) in a standard deck of 52 cards, so $P(B)=\frac{36}{52}=\frac{9}{13}$.

Step3: Calculate P(A')

$P(A') = 1 - P(A)=1-\frac{1}{13}=\frac{12}{13}$.

Step4: Calculate P(B')

$P(B') = 1 - P(B)=1-\frac{9}{13}=\frac{4}{13}$.

Step5: Calculate P(A∩B)

Since an Ace is not a numbered card, $P(A\cap B)=0$.

Step6: Calculate P(A∪B)

Using the formula $P(A\cup B)=P(A)+P(B)-P(A\cap B)$, we have $P(A\cup B)=\frac{1}{13}+\frac{9}{13}-0=\frac{10}{13}$.

Answer:

$P(A)=\frac{1}{13}$, $P(B)=\frac{9}{13}$, $P(A')=\frac{12}{13}$, $P(B')=\frac{4}{13}$, $P(A\cap B)=0$, $P(A\cup B)=\frac{10}{13}$

Explanation:

Step1: Calculate P(A)

The possible combinations for the sum of two dice to be 8 are (2,6), (3,5), (4,4), (5,3), (6,2), so there are 5 combinations. The total number of outcomes when rolling two dice is $n(S)=6\times6 = 36$. So $P(A)=\frac{5}{36}$.

Step2: Calculate P(B)

The number of outcomes where at least one die is a 5: (1,5), (2,5), (3,5), (4,5), (5,5), (6,5), (5,1), (5,2), (5,3), (5,4), (5,6), so there are 11 outcomes. So $P(B)=\frac{11}{36}$.

Step3: Calculate P(A')

$P(A') = 1 - P(A)=1-\frac{5}{36}=\frac{31}{36}$.

Step4: Calculate P(B')

$P(B') = 1 - P(B)=1-\frac{11}{36}=\frac{25}{36}$.

Step5: Calculate P(A∩B)

The combinations that are both a sum of 8 and have at least one 5 are (3,5) and (5,3), so $P(A\cap B)=\frac{2}{36}=\frac{1}{18}$.

Step6: Calculate P(A∪B)

Using the formula $P(A\cup B)=P(A)+P(B)-P(A\cap B)$, we have $P(A\cup B)=\frac{5}{36}+\frac{11}{36}-\frac{2}{36}=\frac{7}{18}$.

Answer:

$P(A)=\frac{5}{36}$, $P(B)=\frac{11}{36}$, $P(A')=\frac{31}{36}$, $P(B')=\frac{25}{36}$, $P(A\cap B)=\frac{1}{18}$, $P(A\cup B)=\frac{7}{18}$

Explanation:

Step1: Calculate P(A)

The possible combinations for the sum of two dice to be 5 or lower are: sum = 2: (1,1); sum = 3: (1,2), (2,1); sum = 4: (1,3), (2,2), (3,1); sum = 5: (1,4), (2,3), (3,2), (4,1). There are 1 + 2+3 + 4=10 combinations. So $P(A)=\frac{10}{36}=\frac{5}{18}$.

Step2: Calculate P(B)

The number of outcomes where the first die is 6 are (6,1), (6,2), (6,3), (6,4), (6,5), (6,6), so $P(B)=\frac{6}{36}=\frac{1}{6}$.

Step3: Calculate P(A')

$P(A') = 1 - P(A)=1-\frac{5}{18}=\frac{13}{18}$.

Step4: Calculate P(B')

$P(B') = 1 - P(B)=1-\frac{1}{6}=\frac{5}{6}$.

Step5: Calculate P(A∩B)

Since if the first die is 6, the sum of two dice is at least 7, $P(A\cap B)=0$.

Step6: Calculate P(A∪B)

Using the formula $P(A\cup B)=P(A)+P(B)-P(A\cap B)$, we have $P(A\cup B)=\frac{5}{18}+\frac{1}{6}-0=\frac{4}{9}$.

Answer:

$P(A)=\frac{5}{18}$, $P(B)=\frac{1}{6}$, $P(A')=\frac{13}{18}$, $P(B')=\frac{5}{6}$, $P(A\cap B)=0$, $P(A\cup B)=\frac{4}{9}$

Explanation:

Step1: Calculate P(A)

If region 1 and region 2 are in A, $P(A)=P(1)+P(2)=0.28 + 0.12=0.4$.

Step2: Calculate P(B)

If region 2 and region 3 are in B, $P(B)=P(2)+P(3)=0.12 + 0.18=0.3$.

Step3: Calculate P(A')

$P(A') = 1 - P(A)=1 - 0.4=0.6$.

Step4: Calculate P(B')

$P(B') = 1 - P(B)=1 - 0.3=0.7$.

Step5: Calculate P(A∩B)

$P(A\cap B)=P(2)=0.12$.

Step6: Calculate P(A∪B)

Using the formula $P(A\cup B)=P(A)+P(B)-P(A\cap B)$, we have $P(A\cup B)=0.4+0.3 - 0.12=0.58$.

Answer:

$P(A)=0.4$, $P(B)=0.3$, $P(A')=0.6$, $P(B')=0.7$, $P(A\cap B)=0.12$, $P(A\cup B)=0.58$

Explanation:

Step1: Calculate P(A)

The total number of people is $24 + 9+9 + 18=60$. The number of females is $24 + 9=33$, so $P(A)=\frac{33}{60}=\frac{11}{20}$.

Step2: Calculate P(B)

The number of people who voted no is $9+18 = 27$, so $P(B)=\frac{27}{60}=\frac{9}{20}$.

Step3: Calculate P(A')

$P(A') = 1 - P(A)=1-\frac{11}{20}=\frac{9}{20}$.

Step4: Calculate P(B')

$P(B') = 1 - P(B)=1-\frac{9}{20}=\frac{11}{20}$.

Step5: Calculate P(A∩B)

The number of females who voted no is 9, so $P(A\cap B)=\frac{9}{60}=\frac{3}{20}$.

Step6: Calculate P(A∪B)

Using the formula $P(A\cup B)=P(A)+P(B)-P(A\cap B)$, we have $P(A\cup B)=\frac{11}{20}+\frac{9}{20}-\frac{3}{20}=\frac{17}{20}$.

Answer:

$P(A)=\frac{11}{20}$, $P(B)=\frac{9}{20}$, $P(A')=\frac{9}{20}$, $P(B')=\frac{11}{20}$, $P(A\cap B)=\frac{3}{20}$, $P(A\cup B)=\frac{17}{20}$