Skip to main content
\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)
Mathematics LibreTexts

18.12: Probability

  • Page ID
    41802
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)

    1. a. \(\frac{6}{13}\) b. \(\frac{2}{13}\) 3. \(\frac{150}{335}=44.8 \%\)

    5. \(\frac{1}{6}\) 7. \(\frac{26}{65}\)

    9. \(\frac{3}{6}=\frac{1}{2}\) 11. \(\frac{4}{52}=\frac{1}{13}\)

    13. \(1-\frac{1}{12}=\frac{11}{12}\) 15. \(1-\frac{25}{65}=\frac{40}{65}\)

    17. \(\frac{1}{6} \cdot \frac{1}{6}=\frac{1}{36}\) 19. \(\frac{1}{6} \cdot \frac{3}{6}=\frac{3}{36}=\frac{1}{12}\)

    21. \(\frac{17}{49} \cdot \frac{16}{48}=\frac{17}{49} \cdot \frac{1}{3}=\frac{17}{147}\)

    23.

    1. \(\frac{4}{52} \cdot \frac{4}{52}=\frac{16}{2704}=\frac{1}{169}\)
    2. \(\frac{4}{52} \cdot \frac{48}{52}=\frac{192}{2704}=\frac{12}{169}\)
    3. \(\frac{48}{52} \cdot \frac{48}{52}=\frac{2304}{2704}=\frac{144}{169}\)
    4. \(\frac{13}{52} \cdot \frac{13}{52}=\frac{169}{2704}=\frac{1}{16}\)
    5. \(\frac{48}{52} \cdot \frac{39}{52}=\frac{1872}{2704}=\frac{117}{169}\)

    25. \(\frac{4}{52} \cdot \frac{4}{51}=\frac{16}{2652}\)

    27.

    1. \(\frac{11}{25} \cdot \frac{14}{24}=\frac{154}{600}\)
    2. \(\frac{14}{25} \cdot \frac{11}{24}=\frac{154}{600}\)
    3. \(\frac{11}{25} \cdot \frac{10}{24}=\frac{110}{600}\)
    4. \(\frac{14}{25} \cdot \frac{13}{24}=\frac{182}{600}\)
    5. no males = two females. Same as part d.

    29. \(P(F \text { and } A)=\frac{10}{65}\)

    31. \(P(\text { red or odd })=\frac{6}{14}+\frac{7}{14}-\frac{3}{14}=\frac{10}{14}\). Or 6 red and 4 odd-numbered blue marbles is 10 out of 14.

    33. \(P(F \text { or } B)=\frac{26}{65}+\frac{22}{65}-\frac{4}{65}=\frac{44}{65}\). Or \(P(F \text { or } B)=\frac{18+4+10+12}{65}=\frac{44}{65}\)

    35. \(\mathrm{P}(\mathrm{King} \text { of Hearts or Queen })=\frac{1}{52}+\frac{4}{52}=\frac{5}{52}\)

    37. a. \(P(\text { even } \mid \mathrm{red})=\frac{2}{5}\) b. \(P(\text { even } \mid \text { red })=\frac{2}{6}\)

    39. \(\mathrm{P}(\text { Heads on second } \mid \text { Tails on first })=\frac{1}{2}\). They are independent events.

    41. \(\mathrm{P}(\text { speak French } \mid \text { female })=\frac{3}{14}\)

    43. Out of 4,000 people, 10 would have the disease. Out of those 10, 9 would test positive, while 1 would falsely test negative. Out of the 3990 uninfected people, 399 would falsely test positive, while 3591 would test negative.

    a. \(P(\text { virus } \mid \text { positive })=\frac{9}{9+399}=\frac{9}{408}=2.2 \%\)

    b. \(\mathrm{P}(\text { no virus } \mid \text { negative })=\frac{3591}{3591+1}=\frac{3591}{3592}=99.97 \%\)

    45. Out of 100,000 people, 300 would have the disease. Of those, 18 would falsely test negative, while 282 would test positive. Of the 99,700 without the disease, 3,988 would falsely test positive and the other 95,712 would test negative.

    \(P(\text { disease } \mid \text { positive })=\frac{282}{282+3988}=\frac{282}{4270}=6.6 \%\)

    47. Out of 100,000 women, 800 would have breast cancer. Out of those, 80 would falsely test negative, while 720 would test positive. Of the 99,200 without cancer, 6,944 would falsely test positive.

    \(P(\text { cancer } \mid \text { positive })=\frac{720}{720+6944}=\frac{720}{7664}=9.4 \%\)

    49. \(2 \cdot 3 \cdot 8 \cdot 2=96\) outfits

    51. a. \(4 \cdot 4 \cdot 4=64\) b. \(4 \cdot 3 \cdot 2=24\)

    53. \(26 \cdot 26 \cdot 26 \cdot 10 \cdot 10 \cdot 10=17,576,000\)

    55. \(_{4} \mathrm{P}_{4}\) or \(4 \cdot 3 \cdot 2 \cdot 1=24\) possible orders

    57. Order matters. \(_7 \mathrm{P}_{4}=840\) possible teams

    59. Order matters. \(_{12} \mathrm{P}_{5}=95,040\) possible themes

    61. Order does not matter. \(_{12} \mathrm{C}_{4}=495\)

    63. \(_{50} \mathrm{C}_{6}=15,890,700\)

    65. \(_{27} \mathrm{C}_{11} \cdot 16=208,606,320\)

    67. There is only 1 way to arrange 5 CD's in alphabetical order. The probability that the CD's are in alphabetical order is one divided by the total number of ways to arrange 5 CD's. Since alphabetical order is only one of all the possible orderings you can either use permutations, or simply use 5!. \(P(\text { alphabetical })=1 / 5 !=1 /(5 P 5)=\frac{1}{120}\).

    69. There are \(_{48} \mathrm{C}_{6}\) total tickets. To match 5 of the 6, a player would need to choose 5 of those 6, \(_6 \mathrm{C}_{5}\), and one of the 42 non-winning numbers, \(_{42} \mathrm{C}_{1}\). \(\frac{6 \cdot 42}{12271512}=\frac{252}{12271512}\)

    71. All possible hands is \(_{52} \mathrm{C}_{5}\)\). Hands will all hearts is \(_{13} \mathrm{C}\)\). \(\frac{1287}{2598960}\).

    73. \(\$ 3\left(\frac{3}{37}\right)+\$ 2\left(\frac{6}{37}\right)+(-\$ 1)\left(\frac{28}{37}\right)=-\$ \frac{7}{37}=-\$ 0.19\)

    75. There are \(_{23} \mathrm{C}_{6}=100,947\)\) possible tickets.

    Expected value \(=\$ 29,999\left(\frac{1}{100947}\right)+(-\$ 1)\left(\frac{100946}{100947}\right)=-\$ 0.70\)

    77. \(\$ 48(0.993)+(-\$ 302)(0.007)=\$ 45.55\)


    18.12: Probability is shared under a CC BY-SA 3.0 license and was authored, remixed, and/or curated by David Lippman via source content that was edited to conform to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.

    • Was this article helpful?