{"id":324,"date":"2026-02-02T19:37:20","date_gmt":"2026-02-02T19:37:20","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/?post_type=chapter&p=324"},"modified":"2026-02-03T18:44:13","modified_gmt":"2026-02-03T18:44:13","slug":"probability-and-counting-theory-get-stronger-answer-key","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/chapter\/probability-and-counting-theory-get-stronger-answer-key\/","title":{"raw":"Probability and Counting Theory: Get Stronger Answer Key","rendered":"Probability and Counting Theory: Get Stronger Answer Key"},"content":{"raw":"

Counting Principles<\/h1>\r\n3.\u00a0The addition principle is applied when determining the total possible of outcomes of either event occurring. The multiplication principle is applied when determining the total possible outcomes of both events occurring. The word \"or\" usually implies an addition problem. The word \"and\" usually implies a multiplication problem.\r\n\r\n7.\u00a0[latex]4+2=6[\/latex]\r\n\r\n9.\u00a0[latex]5+4+7=16[\/latex]\r\n\r\n11.\u00a0[latex]2\\times 6=12[\/latex]\r\n\r\n13.\u00a0[latex]{10}^{3}=1000[\/latex]\r\n\r\n15.\u00a0[latex]P\\left(5,2\\right)=20[\/latex]\r\n\r\n17.\u00a0[latex]P\\left(3,3\\right)=6[\/latex]\r\n\r\n21.\u00a0[latex]C\\left(12,4\\right)=495[\/latex]\r\n\r\n23.\u00a0[latex]C\\left(7,6\\right)=7[\/latex]\r\n\r\n31.\u00a0[latex]\\frac{8!}{3!}=6720[\/latex]\r\n\r\n33.\u00a0[latex]\\frac{12!}{3!2!3!4!}[\/latex]\r\n\r\n35.\u00a09\r\n\r\n37.\u00a0Yes, for the trivial cases [latex]r=0[\/latex] and [latex]r=1[\/latex]. If [latex]r=0[\/latex], then [latex]C\\left(n,r\\right)=P\\left(n,r\\right)=1\\text{.\\hspace{0.17em}}[\/latex] If [latex]r=1[\/latex], then [latex]r=1[\/latex], [latex]C\\left(n,r\\right)=P\\left(n,r\\right)=n[\/latex].\r\n\r\n41.\u00a0[latex]6 - 3+8 - 3=8[\/latex]\r\n\r\n43.\u00a0[latex]4\\times 2\\times 5=40[\/latex]\r\n\r\n45.\u00a0[latex]4\\times 12\\times 3=144[\/latex]\r\n\r\n47.\u00a0[latex]P\\left(15,9\\right)=1,816,214,400[\/latex]\r\n\r\n49.\u00a0[latex]C\\left(10,3\\right)\\times C\\left(6,5\\right)\\times C\\left(5,2\\right)=7,200[\/latex]\r\n\r\n51.\u00a0[latex]{2}^{11}=2048[\/latex]\r\n\r\n53.\u00a0[latex]\\frac{20!}{6!6!8!}=116,396,280[\/latex]\r\n

Binomial Theorem<\/h1>\r\n1.\u00a0A binomial coefficient is an alternative way of denoting the combination [latex]C\\left(n,r\\right)[\/latex]. It is defined as [latex]\\left(\\begin{array}{c}n\\\\ r\\end{array}\\right)=C\\left(n,r\\right)=\\frac{n!}{r!\\left(n-r\\right)!}[\/latex].\r\n\r\n5.\u00a015\r\n\r\n9.\u00a010\r\n\r\n13.\u00a0[latex]64{a}^{3}-48{a}^{2}b+12a{b}^{2}-{b}^{3}[\/latex]\r\n\r\n15.\u00a0[latex]27{a}^{3}+54{a}^{2}b+36a{b}^{2}+8{b}^{3}[\/latex]\r\n\r\n19.\u00a0[latex]1024{x}^{5}-3840{x}^{4}y+5760{x}^{3}{y}^{2}-4320{x}^{2}{y}^{3}+1620x{y}^{4}-243{y}^{5}[\/latex]\r\n\r\n21.\u00a0[latex]\\frac{1}{{x}^{4}}+\\frac{8}{{x}^{3}y}+\\frac{24}{{x}^{2}{y}^{2}}+\\frac{32}{x{y}^{3}}+\\frac{16}{{y}^{4}}[\/latex]\r\n\r\n23.\u00a0[latex]{a}^{17}+17{a}^{16}b+136{a}^{15}{b}^{2}[\/latex]\r\n\r\n25.\u00a0[latex]{a}^{15}-30{a}^{14}b+420{a}^{13}{b}^{2}[\/latex]\r\n\r\n27.\u00a0[latex]3,486,784,401{a}^{20}+23,245,229,340{a}^{19}b+73,609,892,910{a}^{18}{b}^{2}[\/latex]\r\n\r\n29.\u00a0[latex]{x}^{24}-8{x}^{21}\\sqrt{y}+28{x}^{18}y[\/latex]\r\n\r\n33.\u00a0[latex]220,812,466,875,000{y}^{7}[\/latex]\r\n\r\n35.\u00a0[latex]35{x}^{3}{y}^{4}[\/latex]\r\n\r\n39.\u00a0[latex]\\frac{1152{y}^{2}}{{x}^{7}}[\/latex]\r\n\r\n45.\u00a0[latex]590,625{x}^{5}{y}^{2}[\/latex]\r\n\r\n49.\u00a0The expression [latex]{\\left({x}^{3}+2{y}^{2}-z\\right)}^{5}[\/latex] cannot be expanded using the Binomial Theorem because it cannot be rewritten as a binomial.\r\n

Probability<\/h1>\r\n7.\u00a0[latex]\\frac{1}{2}[\/latex].\r\n\r\n9.\u00a0[latex]\\frac{5}{8}[\/latex].\r\n\r\n11.\u00a0[latex]\\frac{1}{2}[\/latex].\r\n\r\n13.\u00a0[latex]\\frac{3}{8}[\/latex].\r\n\r\n14. [latex]S={HH,HT,TH,TT}[\/latex].\r\n\r\n15.\u00a0[latex]\\frac{1}{4}[\/latex].\r\n\r\n16. [latex]\\frac{1}{2}[\/latex].\r\n\r\n17.\u00a0[latex]\\frac{3}{4}[\/latex].\r\n\r\n18. [latex]S={HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, HTTT, THHH, THHT, THTH, THTT, TTHH, TTHT, TTTH, TTTT}[\/latex].\r\n\r\n19.\u00a0[latex]\\frac{3}{8}[\/latex].\r\n\r\n21.\u00a0[latex]\\frac{1}{8}[\/latex].\r\n\r\n23.\u00a0[latex]\\frac{15}{16}[\/latex].\r\n\r\n27.\u00a0[latex]\\frac{1}{13}[\/latex].\r\n\r\n29.\u00a0[latex]\\frac{1}{26}[\/latex].\r\n\r\n31.\u00a0[latex]\\frac{12}{13}[\/latex].\r\n\r\n33.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
<\/th>\r\n1<\/th>\r\n2<\/th>\r\n3<\/th>\r\n4<\/th>\r\n5<\/th>\r\n6<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
1<\/strong><\/td>\r\n(1, 1)\r\n2<\/td>\r\n(1, 2)\r\n3<\/td>\r\n(1, 3)\r\n4<\/td>\r\n(1, 4)\r\n5<\/td>\r\n(1, 5)\r\n6<\/td>\r\n(1, 6)\r\n7<\/td>\r\n<\/tr>\r\n
2<\/strong><\/td>\r\n(2, 1)\r\n3<\/td>\r\n(2, 2)\r\n4<\/td>\r\n(2, 3)\r\n5<\/td>\r\n(2, 4)\r\n6<\/td>\r\n(2, 5)\r\n7<\/td>\r\n(2, 6)\r\n8<\/td>\r\n<\/tr>\r\n
3<\/strong><\/td>\r\n(3, 1)\r\n4<\/td>\r\n(3, 2)\r\n5<\/td>\r\n(3, 3)\r\n6<\/td>\r\n(3, 4)\r\n7<\/td>\r\n(3, 5)\r\n8<\/td>\r\n(3, 6)\r\n9<\/td>\r\n<\/tr>\r\n
4<\/strong><\/td>\r\n(4, 1)\r\n5<\/td>\r\n(4, 2)\r\n6<\/td>\r\n(4, 3)\r\n7<\/td>\r\n(4, 4)\r\n8<\/td>\r\n(4, 5)\r\n9<\/td>\r\n(4, 6)\r\n10<\/td>\r\n<\/tr>\r\n
5<\/strong><\/td>\r\n(5, 1)\r\n6<\/td>\r\n(5, 2)\r\n7<\/td>\r\n(5, 3)\r\n8<\/td>\r\n(5, 4)\r\n9<\/td>\r\n(5, 5)\r\n10<\/td>\r\n(5, 6)\r\n11<\/td>\r\n<\/tr>\r\n
6<\/strong><\/td>\r\n(6, 1)\r\n7<\/td>\r\n(6, 2)\r\n8<\/td>\r\n(6, 3)\r\n9<\/td>\r\n(6, 4)\r\n10<\/td>\r\n(6, 5)\r\n11<\/td>\r\n(6, 6)\r\n12<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n35.\u00a0[latex]\\frac{5}{12}[\/latex].\r\n\r\n37.\u00a0[latex]0[\/latex].\r\n\r\n39.\u00a0[latex]\\frac{4}{9}[\/latex].\r\n\r\n41.\u00a0[latex]\\frac{1}{4}[\/latex].\r\n\r\n43.\u00a0[latex]\\frac{3}{4}[\/latex]\r\n\r\n45.\u00a0[latex]\\frac{21}{26}[\/latex]\r\n\r\n47.\u00a0[latex]\\frac{C\\left(12,5\\right)}{C\\left(48,5\\right)}=\\frac{1}{2162}[\/latex]\r\n\r\n49.\u00a0[latex]\\frac{C\\left(12,3\\right)C\\left(36,2\\right)}{C\\left(48,5\\right)}=\\frac{175}{2162}[\/latex]\r\n\r\n56. [latex]\\frac{40}{317}\\times 100\\%\\approx 12.6\\%[\/latex]\r\n\r\n57.\u00a0[latex]\\frac{C\\left(40000000,1\\right)C\\left(277000000,4\\right)}{C\\left(317000000,5\\right)}=36.78%[\/latex]","rendered":"

Counting Principles<\/h1>\n

3.\u00a0The addition principle is applied when determining the total possible of outcomes of either event occurring. The multiplication principle is applied when determining the total possible outcomes of both events occurring. The word “or” usually implies an addition problem. The word “and” usually implies a multiplication problem.<\/p>\n

7.\u00a0[latex]4+2=6[\/latex]<\/p>\n

9.\u00a0[latex]5+4+7=16[\/latex]<\/p>\n

11.\u00a0[latex]2\\times 6=12[\/latex]<\/p>\n

13.\u00a0[latex]{10}^{3}=1000[\/latex]<\/p>\n

15.\u00a0[latex]P\\left(5,2\\right)=20[\/latex]<\/p>\n

17.\u00a0[latex]P\\left(3,3\\right)=6[\/latex]<\/p>\n

21.\u00a0[latex]C\\left(12,4\\right)=495[\/latex]<\/p>\n

23.\u00a0[latex]C\\left(7,6\\right)=7[\/latex]<\/p>\n

31.\u00a0[latex]\\frac{8!}{3!}=6720[\/latex]<\/p>\n

33.\u00a0[latex]\\frac{12!}{3!2!3!4!}[\/latex]<\/p>\n

35.\u00a09<\/p>\n

37.\u00a0Yes, for the trivial cases [latex]r=0[\/latex] and [latex]r=1[\/latex]. If [latex]r=0[\/latex], then [latex]C\\left(n,r\\right)=P\\left(n,r\\right)=1\\text{.\\hspace{0.17em}}[\/latex] If [latex]r=1[\/latex], then [latex]r=1[\/latex], [latex]C\\left(n,r\\right)=P\\left(n,r\\right)=n[\/latex].<\/p>\n

41.\u00a0[latex]6 - 3+8 - 3=8[\/latex]<\/p>\n

43.\u00a0[latex]4\\times 2\\times 5=40[\/latex]<\/p>\n

45.\u00a0[latex]4\\times 12\\times 3=144[\/latex]<\/p>\n

47.\u00a0[latex]P\\left(15,9\\right)=1,816,214,400[\/latex]<\/p>\n

49.\u00a0[latex]C\\left(10,3\\right)\\times C\\left(6,5\\right)\\times C\\left(5,2\\right)=7,200[\/latex]<\/p>\n

51.\u00a0[latex]{2}^{11}=2048[\/latex]<\/p>\n

53.\u00a0[latex]\\frac{20!}{6!6!8!}=116,396,280[\/latex]<\/p>\n

Binomial Theorem<\/h1>\n

1.\u00a0A binomial coefficient is an alternative way of denoting the combination [latex]C\\left(n,r\\right)[\/latex]. It is defined as [latex]\\left(\\begin{array}{c}n\\\\ r\\end{array}\\right)=C\\left(n,r\\right)=\\frac{n!}{r!\\left(n-r\\right)!}[\/latex].<\/p>\n

5.\u00a015<\/p>\n

9.\u00a010<\/p>\n

13.\u00a0[latex]64{a}^{3}-48{a}^{2}b+12a{b}^{2}-{b}^{3}[\/latex]<\/p>\n

15.\u00a0[latex]27{a}^{3}+54{a}^{2}b+36a{b}^{2}+8{b}^{3}[\/latex]<\/p>\n

19.\u00a0[latex]1024{x}^{5}-3840{x}^{4}y+5760{x}^{3}{y}^{2}-4320{x}^{2}{y}^{3}+1620x{y}^{4}-243{y}^{5}[\/latex]<\/p>\n

21.\u00a0[latex]\\frac{1}{{x}^{4}}+\\frac{8}{{x}^{3}y}+\\frac{24}{{x}^{2}{y}^{2}}+\\frac{32}{x{y}^{3}}+\\frac{16}{{y}^{4}}[\/latex]<\/p>\n

23.\u00a0[latex]{a}^{17}+17{a}^{16}b+136{a}^{15}{b}^{2}[\/latex]<\/p>\n

25.\u00a0[latex]{a}^{15}-30{a}^{14}b+420{a}^{13}{b}^{2}[\/latex]<\/p>\n

27.\u00a0[latex]3,486,784,401{a}^{20}+23,245,229,340{a}^{19}b+73,609,892,910{a}^{18}{b}^{2}[\/latex]<\/p>\n

29.\u00a0[latex]{x}^{24}-8{x}^{21}\\sqrt{y}+28{x}^{18}y[\/latex]<\/p>\n

33.\u00a0[latex]220,812,466,875,000{y}^{7}[\/latex]<\/p>\n

35.\u00a0[latex]35{x}^{3}{y}^{4}[\/latex]<\/p>\n

39.\u00a0[latex]\\frac{1152{y}^{2}}{{x}^{7}}[\/latex]<\/p>\n

45.\u00a0[latex]590,625{x}^{5}{y}^{2}[\/latex]<\/p>\n

49.\u00a0The expression [latex]{\\left({x}^{3}+2{y}^{2}-z\\right)}^{5}[\/latex] cannot be expanded using the Binomial Theorem because it cannot be rewritten as a binomial.<\/p>\n

Probability<\/h1>\n

7.\u00a0[latex]\\frac{1}{2}[\/latex].<\/p>\n

9.\u00a0[latex]\\frac{5}{8}[\/latex].<\/p>\n

11.\u00a0[latex]\\frac{1}{2}[\/latex].<\/p>\n

13.\u00a0[latex]\\frac{3}{8}[\/latex].<\/p>\n

14. [latex]S={HH,HT,TH,TT}[\/latex].<\/p>\n

15.\u00a0[latex]\\frac{1}{4}[\/latex].<\/p>\n

16. [latex]\\frac{1}{2}[\/latex].<\/p>\n

17.\u00a0[latex]\\frac{3}{4}[\/latex].<\/p>\n

18. [latex]S={HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, HTTT, THHH, THHT, THTH, THTT, TTHH, TTHT, TTTH, TTTT}[\/latex].<\/p>\n

19.\u00a0[latex]\\frac{3}{8}[\/latex].<\/p>\n

21.\u00a0[latex]\\frac{1}{8}[\/latex].<\/p>\n

23.\u00a0[latex]\\frac{15}{16}[\/latex].<\/p>\n

27.\u00a0[latex]\\frac{1}{13}[\/latex].<\/p>\n

29.\u00a0[latex]\\frac{1}{26}[\/latex].<\/p>\n

31.\u00a0[latex]\\frac{12}{13}[\/latex].<\/p>\n

33.<\/p>\n\n\n\n\n\n\n\n\n\n\n
<\/th>\n1<\/th>\n2<\/th>\n3<\/th>\n4<\/th>\n5<\/th>\n6<\/th>\n<\/tr>\n<\/thead>\n
1<\/strong><\/td>\n(1, 1)
\n2<\/td>\n		(1, 2)
\n3<\/td>\n		(1, 3)
\n4<\/td>\n		(1, 4)
\n5<\/td>\n		(1, 5)
\n6<\/td>\n		(1, 6)
\n7<\/td>\n<\/tr>\n
2<\/strong><\/td>\n(2, 1)
\n3<\/td>\n		(2, 2)
\n4<\/td>\n		(2, 3)
\n5<\/td>\n		(2, 4)
\n6<\/td>\n		(2, 5)
\n7<\/td>\n		(2, 6)
\n8<\/td>\n<\/tr>\n
3<\/strong><\/td>\n(3, 1)
\n4<\/td>\n		(3, 2)
\n5<\/td>\n		(3, 3)
\n6<\/td>\n		(3, 4)
\n7<\/td>\n		(3, 5)
\n8<\/td>\n		(3, 6)
\n9<\/td>\n<\/tr>\n
4<\/strong><\/td>\n(4, 1)
\n5<\/td>\n		(4, 2)
\n6<\/td>\n		(4, 3)
\n7<\/td>\n		(4, 4)
\n8<\/td>\n		(4, 5)
\n9<\/td>\n		(4, 6)
\n10<\/td>\n<\/tr>\n
5<\/strong><\/td>\n(5, 1)
\n6<\/td>\n		(5, 2)
\n7<\/td>\n		(5, 3)
\n8<\/td>\n		(5, 4)
\n9<\/td>\n		(5, 5)
\n10<\/td>\n		(5, 6)
\n11<\/td>\n<\/tr>\n
6<\/strong><\/td>\n(6, 1)
\n7<\/td>\n		(6, 2)
\n8<\/td>\n		(6, 3)
\n9<\/td>\n		(6, 4)
\n10<\/td>\n		(6, 5)
\n11<\/td>\n		(6, 6)
\n12<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

35.\u00a0[latex]\\frac{5}{12}[\/latex].<\/p>\n

37.\u00a0[latex]0[\/latex].<\/p>\n

39.\u00a0[latex]\\frac{4}{9}[\/latex].<\/p>\n

41.\u00a0[latex]\\frac{1}{4}[\/latex].<\/p>\n

43.\u00a0[latex]\\frac{3}{4}[\/latex]<\/p>\n

45.\u00a0[latex]\\frac{21}{26}[\/latex]<\/p>\n

47.\u00a0[latex]\\frac{C\\left(12,5\\right)}{C\\left(48,5\\right)}=\\frac{1}{2162}[\/latex]<\/p>\n

49.\u00a0[latex]\\frac{C\\left(12,3\\right)C\\left(36,2\\right)}{C\\left(48,5\\right)}=\\frac{175}{2162}[\/latex]<\/p>\n

56. [latex]\\frac{40}{317}\\times 100\\%\\approx 12.6\\%[\/latex]<\/p>\n

57.\u00a0[latex]\\frac{C\\left(40000000,1\\right)C\\left(277000000,4\\right)}{C\\left(317000000,5\\right)}=36.78%[\/latex]<\/p>\n","protected":false},"author":13,"menu_order":12,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":224,"module-header":"- Select Header -","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/pressbooks\/v2\/chapters\/324"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/wp\/v2\/users\/13"}],"version-history":[{"count":3,"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/pressbooks\/v2\/chapters\/324\/revisions"}],"predecessor-version":[{"id":336,"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/pressbooks\/v2\/chapters\/324\/revisions\/336"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/pressbooks\/v2\/parts\/224"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/pressbooks\/v2\/chapters\/324\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/wp\/v2\/media?parent=324"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/pressbooks\/v2\/chapter-type?post=324"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/wp\/v2\/contributor?post=324"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/wp\/v2\/license?post=324"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}