Question

probability rules practice worksheet - fall 2025 - statistics - room 404
for each problem listed below, find p(a), p(b), p(a), p(b), p(a∩b), p(a∪b), and m.e.?
a number is randomly chosen from 1 to 20.

1. a: # greater than 15 b: multiple of 4
2. a: even # b: 11

a single card is randomly chosen from a deck.

1. a: black b: jack
2. a: face b: club
3. a: ace b: number

two dice are rolled.

1. a: sum of 8 b: at least one of the dice is a 5
2. a: sum of 5 or lower is rolled b: the first die rolled is a 6
3. given the following venn diagram and associated probabilities of sample points.

p(1)=.28
p(2)=.12
p(3)=.18
p(4)=.42

1. given the following chart(a: female, b: vote no)

female male
yes 24 9 33
no 9 18 27
33 27 60

Explanation:

Problem 1:

Step1: Find \(P(A)\)

The numbers greater than 15 from 1 - 20 are 16, 17, 18, 19, 20. So \(n(A)=5\). The total number of outcomes \(n = 20\). Then \(P(A)=\frac{n(A)}{n}=\frac{5}{20}=0.25\).

Step2: Find \(P(B)\)

The multiples of 4 from 1 - 20 are 4, 8, 12, 16, 20. So \(n(B)=5\), and \(P(B)=\frac{n(B)}{n}=\frac{5}{20}=0.25\).

Step3: Find \(P(A')\)

\(P(A') = 1 - P(A)=1 - 0.25 = 0.75\).

Step4: Find \(P(B')\)

\(P(B')=1 - P(B)=1 - 0.25 = 0.75\).

Step5: Find \(P(A\cap B)\)

The numbers that are greater than 15 and multiples of 4 are 16, 20. So \(n(A\cap B)=2\), and \(P(A\cap B)=\frac{n(A\cap B)}{n}=\frac{2}{20}=0.1\).

Step6: Find \(P(A\cup B)\)

Using the formula \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\), we have \(P(A\cup B)=0.25 + 0.25- 0.1=0.4\).

Problem 2:

Step1: Find \(P(A)\)

The even numbers from 1 - 20 are 2, 4, 6, 8, 10, 12, 14, 16, 18, 20. So \(n(A) = 10\), and \(P(A)=\frac{n(A)}{n}=\frac{10}{20}=0.5\).

Step2: Find \(P(B)\)

Since there is only one 11 in the set of 1 - 20, \(n(B)=1\), and \(P(B)=\frac{n(B)}{n}=\frac{1}{20}=0.05\).

Step3: Find \(P(A')\)

\(P(A')=1 - P(A)=1 - 0.5 = 0.5\).

Step4: Find \(P(B')\)

\(P(B')=1 - P(B)=1 - 0.05 = 0.95\).

Step5: Find \(P(A\cap B)\)

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

Step6: Find \(P(A\cup B)\)

Using the formula \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\), we get \(P(A\cup B)=0.5 + 0.05-0=0.55\).

Problem 3:

Step1: Find \(P(A)\)

In a standard deck of 52 cards, there are 26 black cards. So \(P(A)=\frac{26}{52}=0.5\).

Step2: Find \(P(B)\)

There are 4 Jacks in a deck, so \(P(B)=\frac{4}{52}=\frac{1}{13}\approx0.077\).

Step3: Find \(P(A')\)

\(P(A')=1 - P(A)=1 - 0.5 = 0.5\).

Step4: Find \(P(B')\)

\(P(B')=1 - P(B)=1-\frac{1}{13}=\frac{12}{13}\approx0.923\).

Step5: Find \(P(A\cap B)\)

There are 2 black Jacks (Jack of spades and Jack of clubs), so \(P(A\cap B)=\frac{2}{52}=\frac{1}{26}\approx0.038\).

Step6: Find \(P(A\cup B)\)

Using the formula \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\), we have \(P(A\cup B)=0.5+\frac{1}{13}-\frac{1}{26}=0.5 + \frac{2 - 1}{26}=0.5+\frac{1}{26}=\frac{13 + 1}{26}=\frac{7}{13}\approx0.538\).

Problem 4:

Step1: Find \(P(A)\)

There are 12 face - cards (4 Jacks, 4 Queens, 4 Kings) in a deck of 52 cards. So \(P(A)=\frac{12}{52}=\frac{3}{13}\approx0.231\).

Step2: Find \(P(B)\)

There are 13 clubs in a deck, so \(P(B)=\frac{13}{52}=0.25\).

Step3: Find \(P(A')\)

\(P(A')=1 - P(A)=1-\frac{3}{13}=\frac{10}{13}\approx0.769\).

Step4: Find \(P(B')\)

\(P(B')=1 - P(B)=1 - 0.25 = 0.75\).

Step5: Find \(P(A\cap B)\)

There are 3 face - cards that are clubs (Jack, Queen, King of clubs), so \(P(A\cap B)=\frac{3}{52}\approx0.058\).

Step6: Find \(P(A\cup B)\)

Using the formula \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\), we get \(P(A\cup B)=\frac{3}{13}+0.25-\frac{3}{52}=\frac{12 + 13 - 3}{52}=\frac{22}{52}=\frac{11}{26}\approx0.423\).

Problem 5:

Step1: Find \(P(A)\)

There are 4 Aces in a deck of 52 cards, so \(P(A)=\frac{4}{52}=\frac{1}{13}\approx0.077\).

Step2: Find \(P(B)\)

There are 36 non - face and non - Ace cards (number cards) in a deck of 52 cards. So \(P(B)=\frac{36}{52}=\frac{9}{13}\approx0.692\).

Step3: Find \(P(A')\)

\(P(A')=1 - P(A)=1-\frac{1}{13}=\frac{12}{13}\approx0.923\).

Step4: Find \(P(B')\)

\(P(B')=1 - P(B)=1-\frac{9}{13}=\frac{4}{13}\approx0.308\).

Step5: Find \(P(A\cap B)\)

Since an Ace is not a number card, \(A\cap B=\varnothing\), so \(P(A\cap B)=0\).

Step6: Find \(P(A\cup B)\)

Using the formula \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\), we have \(P(A\cup B)=\frac{1}{13}+\frac{9}{13}-0=\frac{10}{13}\approx0.769\).

Problem 6:

Step1: Find \(P(A)\)

The possible pairs of two dice that sum to 8 are \((2,6)\), \((3,5)\), \((4,4)\), \((5,3)\), \((6,2)\), so \(n(A)=5\). The total number of outcomes when rolling two dice is \(n = 6\times6 = 36\). So \(P(A)=\frac{5}{36}\approx0.139\).

Step2: Find \(P(B)\)

The pairs where at least one die is a 5 are \((1,5)\), \((2,5)\), \((3,5)\), \((4,5)\), \((5,5)\), \((6,5)\), \((5,1)\), \((5,2)\), \((5,3)\), \((5,4)\), \((5,6)\), so \(n(B)=11\), and \(P(B)=\frac{11}{36}\approx0.306\).

Step3: Find \(P(A')\)

\(P(A')=1 - P(A)=1-\frac{5}{36}=\frac{31}{36}\approx0.861\).

Step4: Find \(P(B')\)

\(P(B')=1 - P(B)=1-\frac{11}{36}=\frac{25}{36}\approx0.694\).

Step5: Find \(P(A\cap B)\)

The pairs that sum to 8 and have at least one 5 are \((3,5)\), \((5,3)\), so \(n(A\cap B)=2\), and \(P(A\cap B)=\frac{2}{36}=\frac{1}{18}\approx0.056\).

Step6: Find \(P(A\cup B)\)

Using the formula \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\), we get \(P(A\cup B)=\frac{5}{36}+\frac{11}{36}-\frac{2}{36}=\frac{14}{36}=\frac{7}{18}\approx0.389\).

Problem 7:

Step1: Find \(P(A)\)

The pairs with a sum of 5 or lower are \((1,1)\), \((1,2)\), \((1,3)\), \((1,4)\), \((2,1)\), \((2,2)\), \((2,3)\), \((3,1)\), \((3,2)\), \((4,1)\), so \(n(A)=10\), and \(P(A)=\frac{10}{36}=\frac{5}{18}\approx0.278\).

Step2: Find \(P(B)\)

The pairs where the first die is 6 are \((6,1)\), \((6,2)\), \((6,3)\), \((6,4)\), \((6,5)\), \((6,6)\), so \(n(B)=6\), and \(P(B)=\frac{6}{36}=\frac{1}{6}\approx0.167\).

Step3: Find \(P(A')\)

\(P(A')=1 - P(A)=1-\frac{5}{18}=\frac{13}{18}\approx0.722\).

Step4: Find \(P(B')\)

\(P(B')=1 - P(B)=1-\frac{1}{6}=\frac{5}{6}\approx0.833\).

Step5: Find \(P(A\cap B)\)

Since the sum of \((6,k)\) for \(k = 1,\cdots,6\) is always greater than 5, \(A\cap B=\varnothing\), so \(P(A\cap B)=0\).

Step6: Find \(P(A\cup B)\)

Using the formula \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\), we have \(P(A\cup B)=\frac{5}{18}+\frac{1}{6}-0=\frac{5 + 3}{18}=\frac{4}{9}\approx0.444\).

Problem 8:

Step1: Find \(P(A)\)

\(P(A)=P(1)+P(2)=0.28 + 0.12=0.4\).

Step2: Find \(P(B)\)

\(P(B)=P(2)+P(3)=0.12 + 0.18=0.3\).

Step3: Find \(P(A')\)

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

Step4: Find \(P(B')\)

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

Step5: Find \(P(A\cap B)\)

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

Step6: Find \(P(A\cup B)\)

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

Problem 9:

Step1: Find \(P(A)\)

\(P(A)=\frac{33}{60}=0.55\).

Step2: Find \(P(B)\)

\(P(B)=\frac{27}{60}=0.45\).

Step3: Find \(P(A')\)

\(P(A')=1 - P(A)=1 - 0.55 = 0.45\).

Step4: Find \(P(B')\)

\(P(B')=1 - P(B)=1 - 0.45 = 0.55\).

Step5: Find \(P(A\cap B)\)

\(P(A\cap B)=\frac{9}{60}=0.15\).

Step6: Find \(P(A\cup B)\)

Using the formula \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\), we have \(P(A\cup B)=0.55 + 0.45-0.15=0.85\).

Answer:

The probabilities for each problem are as calculated above for \(P(A)\), \(P(B)\), \(P(A')\), \(P(B')\), \(P(A\cap B)\), \(P(A\cup B)\) respectively for each sub - problem.