Practice
Rules of Probability Practice
Fifty original questions on equally likely outcomes, replacement, independence, dependence, AND paths, mutually exclusive OR events, and multi-stage probability.
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Question 1
Explanation
There are \(3\) favorable outcomes among \(8\) equally likely outcomes, so \(P=3/8=\frac{3}{8}\).
Question 2
Explanation
There are \(5\) favorable outcomes among \(12\) equally likely outcomes, so \(P=5/12=\frac{5}{12}\).
Question 3
Explanation
There are \(4\) favorable outcomes among \(10\) equally likely outcomes, so \(P=4/10=\frac{2}{5}\).
Question 4
Explanation
There are \(6\) favorable outcomes among \(15\) equally likely outcomes, so \(P=6/15=\frac{2}{5}\).
Question 5
Explanation
There are \(7\) favorable outcomes among \(20\) equally likely outcomes, so \(P=7/20=\frac{7}{20}\).
Question 6
Explanation
Exactly \(2\) of the \(4\) equally likely outcomes satisfy the event, so the probability is \(\frac{1}{2}\).
Question 7
Explanation
Exactly \(2\) of the \(5\) equally likely outcomes satisfy the event, so the probability is \(\frac{2}{5}\).
Question 8
Explanation
Exactly \(1\) of the \(3\) equally likely outcomes satisfy the event, so the probability is \(\frac{1}{3}\).
Question 9
Explanation
Exactly \(2\) of the \(4\) equally likely outcomes satisfy the event, so the probability is \(\frac{1}{2}\).
Question 10
Explanation
Exactly \(3\) of the \(6\) equally likely outcomes satisfy the event, so the probability is \(\frac{1}{2}\).
Question 11
Explanation
Independent is correct because the first result does not change the probabilities for the second result.
Question 12
Explanation
Dependent is correct because the first result changes the probabilities for the second result.
Question 13
Explanation
Independent is correct because the first result does not change the probabilities for the second result.
Question 14
Explanation
Dependent is correct because the first result changes the probabilities for the second result.
Question 15
Explanation
Independent is correct because the first result does not change the probabilities for the second result.
Question 16
Explanation
Replacement restores all \(5\) items. Multiply the independent path: \(\frac{3}{5}\cdot\frac{2}{5}=\frac{6}{25}\).
Question 17
Explanation
Replacement restores all \(9\) items. Multiply the independent path: \(\frac{4}{9}\cdot\frac{5}{9}=\frac{20}{81}\).
Question 18
Explanation
Replacement restores all \(10\) items. Multiply the independent path: \(\frac{6}{10}\cdot\frac{4}{10}=\frac{6}{25}\).
Question 19
Explanation
Replacement restores all \(9\) items. Multiply the independent path: \(\frac{2}{9}\cdot\frac{7}{9}=\frac{14}{81}\).
Question 20
Explanation
Replacement restores all \(8\) items. Multiply the independent path: \(\frac{5}{8}\cdot\frac{3}{8}=\frac{15}{64}\).
Question 21
Explanation
The first probability is \(5/8\). Then \(7\) items remain and \(3\) satisfy the second event, giving \(\frac{15}{56}\).
Question 22
Explanation
The first probability is \(4/10\). Then \(9\) items remain and \(3\) satisfy the second event, giving \(\frac{2}{15}\).
Question 23
Explanation
The first probability is \(7/9\). Then \(8\) items remain and \(2\) satisfy the second event, giving \(\frac{7}{36}\).
Question 24
Explanation
The first probability is \(3/8\). Then \(7\) items remain and \(2\) satisfy the second event, giving \(\frac{3}{28}\).
Question 25
Explanation
The first probability is \(6/10\). Then \(9\) items remain and \(4\) satisfy the second event, giving \(\frac{4}{15}\).
Question 26
Explanation
Because the events cannot overlap, add their favorable counts: \(3+2=5\). Thus the probability is \(\frac{5}{12}\).
Question 27
Explanation
Because the events cannot overlap, add their favorable counts: \(1+4=5\). Thus the probability is \(\frac{1}{2}\).
Question 28
Explanation
Because the events cannot overlap, add their favorable counts: \(5+3=8\). Thus the probability is \(\frac{8}{15}\).
Question 29
Explanation
Because the events cannot overlap, add their favorable counts: \(6+7=13\). Thus the probability is \(\frac{13}{20}\).
Question 30
Explanation
Because the events cannot overlap, add their favorable counts: \(4+5=9\). Thus the probability is \(\frac{1}{2}\).
Question 31
Explanation
Add the two valid branches: \(P(AA)+P(BB)=\frac{4(3)+3(2)}{7(6)}=\frac{3}{7}\).
Question 32
Explanation
Add the two valid branches: \(P(AA)+P(BB)=\frac{5(5)+2(2)}{7(7)}=\frac{29}{49}\).
Question 33
Explanation
Add the two valid branches: \(P(AA)+P(BB)=\frac{6(5)+4(3)}{10(9)}=\frac{7}{15}\).
Question 34
Explanation
Add the two valid branches: \(P(AA)+P(BB)=\frac{3(3)+5(5)}{8(8)}=\frac{17}{32}\).
Question 35
Explanation
Add the two valid branches: \(P(AA)+P(BB)=\frac{7(6)+3(2)}{10(9)}=\frac{8}{15}\).
Question 36
Explanation
With replacement the probability is 0.1406; without replacement it is 0.1071. Therefore, with replacement gives the greater value.
Question 37
Explanation
With replacement the probability is 0.2400; without replacement it is 0.2667. Therefore, without replacement gives the greater value.
Question 38
Explanation
With replacement the probability is 0.6049; without replacement it is 0.5833. Therefore, with replacement gives the greater value.
Question 39
Explanation
With replacement the probability is 0.2469; without replacement it is 0.2778. Therefore, without replacement gives the greater value.
Question 40
Explanation
With replacement the probability is 0.4444; without replacement it is 0.4167. Therefore, with replacement gives the greater value.
Question 41
Explanation
Multiply the three path probabilities with the stated replacement rule: \(\frac{3}{5}\cdot\frac{2}{5}\cdot\frac{3}{5}=\frac{18}{125}\).
Question 42
Explanation
Multiply the three path probabilities with the stated replacement rule: \(\frac{5}{8}\cdot\frac{4}{7}\cdot\frac{3}{6}=\frac{5}{28}\).
Question 43
Explanation
Multiply the three path probabilities with the stated replacement rule: \(\frac{6}{10}\cdot\frac{4}{10}\cdot\frac{6}{10}=\frac{18}{125}\).
Question 44
Explanation
Multiply the three path probabilities with the stated replacement rule: \(\frac{6}{8}\cdot\frac{2}{7}\cdot\frac{1}{6}=\frac{1}{28}\).
Question 45
Explanation
Multiply the three path probabilities with the stated replacement rule: \(\frac{4}{7}\cdot\frac{3}{6}\cdot\frac{3}{5}=\frac{6}{35}\).
Question 46
Explanation
List the favorable integers and count overlapping values only once. There are \(9\) distinct favorable outcomes among \(18\), giving \(\frac{1}{2}\).
Question 47
Explanation
List the favorable integers and count overlapping values only once. There are \(8\) distinct favorable outcomes among \(20\), giving \(\frac{2}{5}\).
Question 48
Explanation
List the favorable integers and count overlapping values only once. There are \(9\) distinct favorable outcomes among \(15\), giving \(\frac{3}{5}\).
Question 49
Explanation
List the favorable integers and count overlapping values only once. There are \(10\) distinct favorable outcomes among \(16\), giving \(\frac{5}{8}\).
Question 50
Explanation
List the favorable integers and count overlapping values only once. There are \(11\) distinct favorable outcomes among \(12\), giving \(\frac{11}{12}\).
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Questions to review
No mistakes this time. Excellent work.
- Question 1Basic probabilityEasy
- Question 2Basic probabilityEasy
- Question 3Basic probabilityEasy
- Question 4Basic probabilityEasy
- Question 5Basic probabilityEasy
- Question 6Sample-space reasoningEasy
- Question 7Sample-space reasoningEasy
- Question 8Sample-space reasoningEasy
- Question 9Sample-space reasoningEasy
- Question 10Sample-space reasoningEasy
- Question 11Independent and dependent eventsEasy
- Question 12Independent and dependent eventsEasy
- Question 13Independent and dependent eventsEasy
- Question 14Independent and dependent eventsEasy
- Question 15Independent and dependent eventsEasy
- Question 16Independent probability with replacementMedium
- Question 17Independent probability with replacementMedium
- Question 18Independent probability with replacementMedium
- Question 19Independent probability with replacementMedium
- Question 20Independent probability with replacementMedium
- Question 21Dependent probability without replacementMedium
- Question 22Dependent probability without replacementMedium
- Question 23Dependent probability without replacementMedium
- Question 24Dependent probability without replacementMedium
- Question 25Dependent probability without replacementMedium
- Question 26Mutually exclusive OR probabilityMedium
- Question 27Mutually exclusive OR probabilityMedium
- Question 28Mutually exclusive OR probabilityMedium
- Question 29Mutually exclusive OR probabilityMedium
- Question 30Mutually exclusive OR probabilityMedium
- Question 31Same-category probabilityMedium
- Question 32Same-category probabilityMedium
- Question 33Same-category probabilityMedium
- Question 34Same-category probabilityMedium
- Question 35Same-category probabilityMedium
- Question 36Compare replacement conditionsMedium
- Question 37Compare replacement conditionsMedium
- Question 38Compare replacement conditionsMedium
- Question 39Compare replacement conditionsMedium
- Question 40Compare replacement conditionsMedium
- Question 41Multi-stage probabilityHard
- Question 42Multi-stage probabilityHard
- Question 43Multi-stage probabilityHard
- Question 44Multi-stage probabilityHard
- Question 45Multi-stage probabilityHard
- Question 46Overlapping OR by direct countingHard
- Question 47Overlapping OR by direct countingHard
- Question 48Overlapping OR by direct countingHard
- Question 49Overlapping OR by direct countingHard
- Question 50Overlapping OR by direct countingHard