statistics sequential probability worksheet 2020\n\in a row\ vocabularies\nfor coin flips:\n1. p(2 heads in…

statistics sequential probability worksheet 2020\n\in a row\ vocabularies\nfor coin flips:\n1. p(2 heads in a row)\n2. p(3 tails in a row)\n3. p(5 heads in a row)\n4. p(n heads in a row)\nfor single - die rolls:\n5. p(2 sixes in a row)\n6. p(3 fives in a row)\n7. p(4 non - twos in a row)\n8. p(n threes in a row)\nsequential probabilities\nfor single die rolls:\n9. p(three, five)\n10. p(four, non - six)\n11. p(even, multiple of 3)\n12. p(one, odd, multiple of 3)\n13. p(non - four, non - five, two)\nfor a deck of cards (calculate with replacement, then without replacement):\n14. p(ace, king)\n15. p(diamond, red)\n16. p(face, ace, ten)\n17. p(ace, king, queen, jack, ten)\nfor a bag of 10 marbles (5 red, 3 white, 1 blue)(calculate with and without replacement):\n18. p(red, red)\n19. p(white, white)\n20. p(blue, blue)\n21. p(red, blue)\n22. p(white, non - blue)\n23. p(red, white, blue)\n24. p(white, green)\n25. p(white, white, white)

statistics sequential probability worksheet 2020\n\in a row\ vocabularies\nfor coin flips:\n1. p(2 heads in a row)\n2. p(3 tails in a row)\n3. p(5 heads in a row)\n4. p(n heads in a row)\nfor single - die rolls:\n5. p(2 sixes in a row)\n6. p(3 fives in a row)\n7. p(4 non - twos in a row)\n8. p(n threes in a row)\nsequential probabilities\nfor single die rolls:\n9. p(three, five)\n10. p(four, non - six)\n11. p(even, multiple of 3)\n12. p(one, odd, multiple of 3)\n13. p(non - four, non - five, two)\nfor a deck of cards (calculate with replacement, then without replacement):\n14. p(ace, king)\n15. p(diamond, red)\n16. p(face, ace, ten)\n17. p(ace, king, queen, jack, ten)\nfor a bag of 10 marbles (5 red, 3 white, 1 blue)(calculate with and without replacement):\n18. p(red, red)\n19. p(white, white)\n20. p(blue, blue)\n21. p(red, blue)\n22. p(white, non - blue)\n23. p(red, white, blue)\n24. p(white, green)\n25. p(white, white, white)

Answer

Explanation:

Step1: Recall probability formula for independent events

For independent events, if the probability of event (A) is (P(A)) and event (B) is (P(B)), the probability of (A) and (B) occurring in sequence is (P(A)\times P(B)). For a coin - flip, (P(\text{head})=P(\text{tail})=\frac{1}{2}), for a die - roll (P(i)=\frac{1}{6}) where (i = 1,2,\cdots,6), for a standard deck of 52 cards, (P(\text{ace})=\frac{4}{52}=\frac{1}{13}), (P(\text{king})=\frac{4}{52}=\frac{1}{13}), (P(\text{diamond})=\frac{13}{52}=\frac{1}{4}), (P(\text{red})=\frac{26}{52}=\frac{1}{2}), (P(\text{face})=\frac{12}{52}=\frac{3}{13}), (P(\text{ten})=\frac{4}{52}=\frac{1}{13}), and for marbles in a bag, if there are (n) marbles in total and (n_i) marbles of a certain type, (P(\text{type }i)=\frac{n_i}{n}).

Step2: Calculate probabilities for coin - flips

  1. (P(2\text{ heads in a row})): Since the flips are independent, (P(\text{head})\times P(\text{head})=\frac{1}{2}\times\frac{1}{2}=\frac{1}{4})
  2. (P(3\text{ tails in a row})): (P(\text{tail})\times P(\text{tail})\times P(\text{tail})=\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}=\frac{1}{8})
  3. (P(5\text{ heads in a row})): (P(\text{head})^5=\left(\frac{1}{2}\right)^5=\frac{1}{32})
  4. (P(n\text{ heads in a row})): (P(\text{head})^n=\left(\frac{1}{2}\right)^n)

Step3: Calculate probabilities for single - die rolls

  1. (P(2\text{ sixes in a row})): (P(\text{six})\times P(\text{six})=\frac{1}{6}\times\frac{1}{6}=\frac{1}{36})
  2. (P(3\text{ fives in a row})): (P(\text{five})^3=\left(\frac{1}{6}\right)^3=\frac{1}{216})
  3. (P(4\text{ non - twos in a row})): (P(\text{non - two})=\frac{5}{6}), so (P(\text{non - two})^4=\left(\frac{5}{6}\right)^4=\frac{625}{1296})
  4. (P(n\text{ threes in a row})): (P(\text{three})^n=\left(\frac{1}{6}\right)^n)

Step4: Calculate sequential probabilities for single - die rolls

  1. (P(\text{three,five})): (P(\text{three})\times P(\text{five})=\frac{1}{6}\times\frac{1}{6}=\frac{1}{36})
  2. (P(\text{four,non - six})): (P(\text{four})\times P(\text{non - six})=\frac{1}{6}\times\frac{5}{6}=\frac{5}{36})
  3. (P(\text{even,multiple of }3)): (P(\text{even})=\frac{3}{6}), (P(\text{multiple of }3)=\frac{2}{6}), (P(\text{even,multiple of }3)=\frac{3}{6}\times\frac{2}{6}=\frac{1}{6})
  4. (P(\text{one,odd,multiple of }3)): (P(\text{one})\times P(\text{odd})\times P(\text{multiple of }3)=\frac{1}{6}\times\frac{3}{6}\times\frac{2}{6}=\frac{1}{36})
  5. (P(\text{non - four,non - five,two})): (P(\text{non - four})=\frac{5}{6}), (P(\text{non - five})=\frac{5}{6}), (P(\text{two})=\frac{1}{6}), (P=\frac{5}{6}\times\frac{5}{6}\times\frac{1}{6}=\frac{25}{216})

Step5: Calculate probabilities for card draws

With replacement

  1. (P(\text{Ace,King})): (P(\text{Ace})\times P(\text{King})=\frac{4}{52}\times\frac{4}{52}=\frac{1}{169})
  2. (P(\text{Diamond,Red})): (P(\text{Diamond})=\frac{13}{52}), (P(\text{Red})=\frac{26}{52}), (P=\frac{13}{52}\times\frac{26}{52}=\frac{1}{8})
  3. (P(\text{Face,Ace,Ten})): (P(\text{Face})\times P(\text{Ace})\times P(\text{Ten})=\frac{12}{52}\times\frac{4}{52}\times\frac{4}{52}=\frac{12}{2197})
  4. (P(\text{Ace,King,Queen,Jack,Ten})): (P(\text{Ace})\times P(\text{King})\times P(\text{Queen})\times P(\text{Jack})\times P(\text{Ten})=\frac{4}{52}\times\frac{4}{52}\times\frac{4}{52}\times\frac{4}{52}\times\frac{4}{52}=\frac{1024}{380204032})

Without replacement

  1. (P(\text{Ace,King})): (P(\text{Ace})=\frac{4}{52}), (P(\text{King}|\text{Ace})=\frac{4}{51}), (P=\frac{4}{52}\times\frac{4}{51}=\frac{4}{663})
  2. (P(\text{Diamond,Red})): (P(\text{Diamond})=\frac{13}{52}), (P(\text{Red}|\text{Diamond})=\frac{25}{51}), (P=\frac{13}{52}\times\frac{25}{51}=\frac{25}{204})
  3. (P(\text{Face,Ace,Ten})): (P(\text{Face})=\frac{12}{52}), (P(\text{Ace}|\text{Face})=\frac{4}{51}), (P(\text{Ten}|\text{Face,Ace})=\frac{4}{50}), (P=\frac{12}{52}\times\frac{4}{51}\times\frac{4}{50}=\frac{8}{5525})
  4. (P(\text{Ace,King,Queen,Jack,Ten})): (P(\text{Ace})=\frac{4}{52}), (P(\text{King}|\text{Ace})=\frac{4}{51}), (P(\text{Queen}|\text{Ace,King})=\frac{4}{50}), (P(\text{Jack}|\text{Ace,King,Queen})=\frac{4}{49}), (P(\text{Ten}|\text{Ace,King,Queen,Jack})=\frac{4}{48}), (P=\frac{4\times4\times4\times4\times4}{52\times51\times50\times49\times48}=\frac{4}{31187520})

Step6: Calculate probabilities for marbles

With replacement

  1. (P(\text{Red,Red})): (P(\text{Red})=\frac{6}{10}), (P=\frac{6}{10}\times\frac{6}{10}=\frac{9}{25})
  2. (P(\text{White,White})): (P(\text{White})=\frac{3}{10}), (P=\frac{3}{10}\times\frac{3}{10}=\frac{9}{100})
  3. (P(\text{Blue,Blue})): (P(\text{Blue})=\frac{1}{10}), (P=\frac{1}{10}\times\frac{1}{10}=\frac{1}{100})
  4. (P(\text{Red,Blue})): (P(\text{Red})=\frac{6}{10}), (P(\text{Blue})=\frac{1}{10}), (P=\frac{6}{10}\times\frac{1}{10}=\frac{3}{50})
  5. (P(\text{White,non - Blue})): (P(\text{White})=\frac{3}{10}), (P(\text{non - Blue})=\frac{9}{10}), (P=\frac{3}{10}\times\frac{9}{10}=\frac{27}{100})
  6. (P(\text{Red,White,Blue})): (P(\text{Red})\times P(\text{White})\times P(\text{Blue})=\frac{6}{10}\times\frac{3}{10}\times\frac{1}{10}=\frac{9}{500})
  7. (P(\text{White,Green})): Since there are no green marbles, (P = 0)
  8. (P(\text{White,White,White})): (P(\text{White})^3=\left(\frac{3}{10}\right)^3=\frac{27}{1000})

Without replacement

  1. (P(\text{Red,Red})): (P(\text{Red})=\frac{6}{10}), (P(\text{Red}|\text{Red})=\frac{5}{9}), (P=\frac{6}{10}\times\frac{5}{9}=\frac{1}{3})
  2. (P(\text{White,White})): (P(\text{White})=\frac{3}{10}), (P(\text{White}|\text{White})=\frac{2}{9}), (P=\frac{3}{10}\times\frac{2}{9}=\frac{1}{15})
  3. (P(\text{Blue,Blue})): Since there is only 1 blue marble, (P = 0)
  4. (P(\text{Red,Blue})): (P(\text{Red})=\frac{6}{10}), (P(\text{Blue}|\text{Red})=\frac{1}{9}), (P=\frac{6}{10}\times\frac{1}{9}=\frac{1}{15})
  5. (P(\text{White,non - Blue})): (P(\text{White})=\frac{3}{10}), (P(\text{non - Blue}|\text{White})=\frac{9}{9} = 1), (P=\frac{3}{10}\times1=\frac{3}{10})
  6. (P(\text{Red,White,Blue})): (P(\text{Red})=\frac{6}{10}), (P(\text{White}|\text{Red})=\frac{3}{9}), (P(\text{Blue}|\text{Red,White})=\frac{1}{8}), (P=\frac{6}{10}\times\frac{3}{9}\times\frac{1}{8}=\frac{1}{40})
  7. (P(\text{White,Green})): Since there are no green marbles, (P = 0)
  8. (P(\text{White,White,White})): (P(\text{White})=\frac{3}{10}), (P(\text{White}|\text{White})=\frac{2}{9}), (P(\text{White}|\text{White,White})=\frac{1}{8}), (P=\frac{3}{10}\times\frac{2}{9}\times\frac{1}{8}=\frac{1}{120})

Answer:

  1. (\frac{1}{4})
  2. (\frac{1}{8})
  3. (\frac{1}{32})
  4. (\left(\frac{1}{2}\right)^n)
  5. (\frac{1}{36})
  6. (\frac{1}{216})
  7. (\frac{625}{1296})
  8. (\left(\frac{1}{6}\right)^n)
  9. (\frac{1}{36})
  10. (\frac{5}{36})
  11. (\frac{1}{6})
  12. (\frac{1}{36})
  13. (\frac{25}{216})
  14. With replacement: (\frac{1}{169}), Without replacement: (\frac{4}{663})
  15. With replacement: (\frac{1}{8}), Without replacement: (\frac{25}{204})
  16. With replacement: (\frac{12}{2197}), Without replacement: (\frac{8}{5525})
  17. With replacement: (\frac{1024}{380204032}), Without replacement: (\frac{4}{31187520})
  18. With replacement: (\frac{9}{25}), Without replacement: (\frac{1}{3})
  19. With replacement: (\frac{9}{100}), Without replacement: (\frac{1}{15})
  20. With replacement: (\frac{1}{100}), Without replacement: (0)
  21. With replacement: (\frac{3}{50}), Without replacement: (\frac{1}{15})
  22. With replacement: (\frac{27}{100}), Without replacement: (\frac{3}{10})
  23. With replacement: (\frac{9}{500}), Without replacement: (\frac{1}{40})
  24. (0)
  25. With replacement: (\frac{27}{1000}), Without replacement: (\frac{1}{120})