{"id":68139,"date":"2021-08-30T16:24:00","date_gmt":"2021-08-30T10:54:00","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=68139"},"modified":"2021-08-30T16:24:25","modified_gmt":"2021-08-30T10:54:25","slug":"rd-sharma-solutions-class-11-maths-chapter-33-exercise-33-4","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-33-exercise-33-4\/","title":{"rendered":"RD Sharma Class 11 Solutions Chapter 33 Probability Exercise 33.4 (Updated for 2021-22)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-121471\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Class-11-Solutions-Chapter-33-Exercise-33.4.jpg\" alt=\"RD Sharma Solutions Class 11 Maths Chapter 33 Exercise 33.4\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Class-11-Solutions-Chapter-33-Exercise-33.4.jpg 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Class-11-Solutions-Chapter-33-Exercise-33.4-768x432.jpg 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><strong>RD Sharma Solutions Class 11 Maths Chapter 33 Exercise 33.4:\u00a0<\/strong>Until now, we&#8217;ve computed the likelihood of an event occurring or not occurring using the favourable and total number of elementary events. The <a href=\"https:\/\/www.kopykitab.com\/blog\/cbse-class-11-maths-rd-sharma-solutions\/\"><strong>RD Sharma Class 11 Solutions<\/strong><\/a> are described in simple language by topic experts to assist students in improving their analytical thinking and problem-solving abilities. <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-33-probability\/\"><strong>RD Sharma Solutions Class 11 Maths Chapter 33<\/strong><\/a> Exercise 33.4\u00a0pdf can be downloaded for free to help students improve their exam preparation and score higher on their board exams.<\/p>\n<div id=\"ez-toc-container\" class=\"ez-toc-v2_0_47_1 counter-hierarchy ez-toc-counter ez-toc-grey ez-toc-container-direction\">\n<div class=\"ez-toc-title-container\">\n<p class=\"ez-toc-title\">Table of Contents<\/p>\n<span class=\"ez-toc-title-toggle\"><a href=\"#\" class=\"ez-toc-pull-right ez-toc-btn ez-toc-btn-xs ez-toc-btn-default ez-toc-toggle\" 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Sharma Class 11 Solutions Chapter 33 Statistics Exercise 33.4?\">How many questions are there in RD Sharma Class 11 Solutions Chapter 33 Statistics Exercise 33.4?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-5\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-33-exercise-33-4\/#what-are-the-advantages-of-using-rd-sharma-solutions-class-11-maths-chapter-33-exercise-334\" title=\"What are the advantages of using RD Sharma Solutions Class 11 Maths Chapter 33 Exercise 33.4?\">What are the advantages of using RD Sharma Solutions Class 11 Maths Chapter 33 Exercise 33.4?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-6\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-33-exercise-33-4\/#where-can-i-download-rd-sharma-solutions-class-11-maths-chapter-33-exercise-334-free-pdf\" title=\"Where can I download RD Sharma Solutions Class 11 Maths Chapter 33 Exercise 33.4 free PDF?\">Where can I download RD Sharma Solutions Class 11 Maths Chapter 33 Exercise 33.4 free PDF?<\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"download-rd-sharma-class-11-solutions-chapter-33-probability-exercise-334-free-pdf\"><\/span>Download RD Sharma Class 11 Solutions Chapter 33 Probability Exercise 33.4 Free PDF<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p><a href=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Solutions-Class-11-Maths-Chapter-33-Ex-33.4-1.pdf\">RD Sharma Solutions Class 11 Maths Chapter 33 Exercise 33.4<\/a><\/p>\n<div id=\"example1\" style=\"text-align: justify;\">\u00a0<\/div>\n<p style=\"text-align: justify;\"><style>\n.pdfobject-container { height: 800px;}<br \/>\n.pdfobject { border: 1px solid #666; }<br \/>\n<\/style><\/p>\n<p style=\"text-align: justify;\"><script src=\"https:\/\/www.kopykitab.com\/_utility\/js\/pdfobject.min.js\"><\/script><br \/><script>PDFObject.embed(\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Solutions-Class-11-Maths-Chapter-33-Ex-33.4-1.pdf\", \"#example1\");<\/script><\/p>\n<h3><span class=\"ez-toc-section\" id=\"access-rd-sharma-solutions-class-11-maths-chapter-33-exercise-334-important-question-with-answers\"><\/span>Access RD Sharma Solutions Class 11 Maths Chapter 33 Exercise 33.4- Important Question with Answers<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>1. (a)<\/strong>\u00a0<strong>If A and B be mutually exclusive events associated with a random experiment such that P (A) = 0.4 and P (B) = 0.5, then find:<br \/>(i) P(A\u00a0\u222a\u00a0B)<\/strong><\/p>\n<p><img class=\"alignnone size-full wp-image-121516\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/a-1.png\" alt=\"a\" width=\"113\" height=\"82\" \/><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given: A and B are two mutually exclusive events.<\/p>\n<p>P (A) = 0.4 and P (B) = 0.5<\/p>\n<p>By definition of mutually exclusive events we know that:<\/p>\n<p>P (A\u00a0\u222a\u00a0B) = P (A) + P (B)<\/p>\n<p><br \/>Now, we have to find<\/p>\n<p><strong>(i)<\/strong>\u00a0P (A\u00a0\u222a\u00a0B) = P (A) + P (B) = 0.5 + 0.4 = 0.9<\/p>\n<p><br \/><strong>(ii)<\/strong>\u00a0P (A\u2032\u00a0\u2229\u00a0B\u2032) = P (A\u00a0\u222a\u00a0B)\u2032\u00a0{using De Morgan\u2019s Law}<\/p>\n<p>P (A\u2032\u00a0\u2229\u00a0B\u2032) = 1\u00a0\u2013\u00a0P (A\u00a0\u222a\u00a0B)<\/p>\n<p>= 1 \u2013 0.9<\/p>\n<p>= 0.1<\/p>\n<p><br \/><strong>(iii)<\/strong>\u00a0P (A\u2032\u00a0\u2229\u00a0B) [This indicates only the part which is common with B and not A.<\/p>\n<p>Hence this indicates only B]<br \/>P (only B) = P (B) \u2013 P (A\u00a0\u2229\u00a0B)<\/p>\n<p>As A and B are mutually exclusive so they don\u2019t have any common parts.<\/p>\n<p>P (A\u00a0\u2229\u00a0B) = 0<\/p>\n<p>\u2234\u00a0P (A\u2032\u00a0\u2229\u00a0B) = P (B) = 0.5<\/p>\n<p><br \/><strong>(iv)<\/strong>\u00a0P (A\u00a0\u2229\u00a0B\u2032) [This indicates only the part which is common with A and not B.<\/p>\n<p>Hence this indicates only A]<\/p>\n<p>P (only A) = P (A) \u2013 P (A\u00a0\u2229\u00a0B)<\/p>\n<p>As A and B are mutually exclusive so they don\u2019t have any common parts.<\/p>\n<p>P (A\u00a0\u2229\u00a0B) = 0<\/p>\n<p>\u2234\u00a0P (A\u00a0\u2229\u00a0B\u2032) = P (A) = 0.4<\/p>\n<p><strong>(b)<\/strong>\u00a0<strong>A and B are two events such that P (A) = 0.54, P (B) = 0.69 and P (A \u2229 B) = 0.35. Find (i) <\/strong><strong>P (A\u00a0\u222a\u00a0B)<\/strong><\/p>\n<p><img class=\"alignnone size-full wp-image-121517\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/b-2.png\" alt=\"b\" width=\"115\" height=\"90\" \/><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given: A and B are two events.<\/p>\n<p>P (A) = 0.54, P (B) = 0.69 and P (A\u00a0\u2229\u00a0B) = 0.35<\/p>\n<p>By definition of P (A or B) under the axiomatic approach we know that:<\/p>\n<p>P (A\u00a0\u222a\u00a0B) = P (A) + P (B) \u2013 P (A\u00a0\u2229\u00a0B)<\/p>\n<p><br \/>Now we have to find:<\/p>\n<p><strong>(i)<\/strong>\u00a0P (A\u00a0\u222a\u00a0B) = P (A) + P (B) \u2013 P (A\u00a0\u2229\u00a0B)<\/p>\n<p>= 0.54 + 0.69 \u2013 0.35<\/p>\n<p>= 0.88<\/p>\n<p><strong>(ii)<\/strong>\u00a0P (A\u2032\u00a0\u2229\u00a0B\u2032) = P (A\u00a0\u222a\u00a0B)\u2032\u00a0{using De Morgan\u2019s Law}<\/p>\n<p>P (A\u2032\u00a0\u2229\u00a0B\u2032) = 1\u00a0\u2013\u00a0P (A\u00a0\u222a\u00a0B)<\/p>\n<p>= 1 \u2013 0.88<\/p>\n<p>= 0.12<\/p>\n<p><br \/><strong>(iii)<\/strong>\u00a0P (A\u00a0\u2229\u00a0B\u2032) [This indicates only the part which is common with A and not B.<\/p>\n<p>Hence this indicates only A]<\/p>\n<p>P (only A) = P (A) \u2013 P (A\u00a0\u2229\u00a0B)<\/p>\n<p>\u2234\u00a0P (A\u00a0\u2229\u00a0B\u2032) = P (A) \u2013 P (A\u00a0\u2229\u00a0B)<\/p>\n<p>= 0.54 \u2013 0.35<\/p>\n<p>= 0.19<\/p>\n<p><br \/><strong>(iv)<\/strong>\u00a0P (A\u2032\u00a0\u2229\u00a0B) [This indicates only the part which is common with B and not A.<\/p>\n<p>Hence this indicates only B]<\/p>\n<p>P (only B) = P (B) \u2013 P (A\u00a0\u2229\u00a0B)<\/p>\n<p>\u2234\u00a0P (A\u2032\u00a0\u2229\u00a0B) = P (B) \u2013 P (A\u00a0\u2229\u00a0B)<\/p>\n<p>= 0.69 \u2013 0.35<\/p>\n<p>= 0.34<\/p>\n<p><strong>(c)<\/strong>\u00a0<strong>Fill in the blanks in the following table:<\/strong><\/p>\n<p><img class=\"alignnone size-full wp-image-121518\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/c-1.png\" alt=\"c\" width=\"345\" height=\"115\" \/><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><strong>(i)<\/strong>\u00a0By definition of P (A or B) under axiomatic approach we know that:<\/p>\n<p>P (A\u00a0\u222a\u00a0B) = P (A) + P (B) \u2013 P (A\u00a0\u2229\u00a0B)<\/p>\n<p>By using data from table, we get:<\/p>\n<p>\u2234\u00a0P (A\u00a0\u222a\u00a0B) = 1\/3 + 1\/5 \u2013 1\/15<\/p>\n<p>= 8\/15 \u2013 1\/15<\/p>\n<p>= 7\/15<\/p>\n<p><strong>(ii)<\/strong>\u00a0By definition of P (A or B) under axiomatic approach we know that:<\/p>\n<p>P (A\u00a0\u222a\u00a0B) = P (A) + P (B) \u2013 P (A\u00a0\u2229\u00a0B)<\/p>\n<p>P (B) = P (A\u00a0\u222a\u00a0B) + P (A\u00a0\u2229\u00a0B) \u2013 P (A)<\/p>\n<p>By using data from table, we get:<\/p>\n<p>\u2234\u00a0P (B) = 0.6 + 0.25 \u2013 0.35<\/p>\n<p>= 0.5<\/p>\n<p><br \/><strong>(iii)<\/strong>\u00a0By definition of P (A or B) under axiomatic approach we know that:<\/p>\n<p>P (A\u00a0\u222a\u00a0B) = P (A) + P (B) \u2013 P (A\u00a0\u2229\u00a0B)<\/p>\n<p>P (A\u00a0\u2229\u00a0B)\u00a0=\u00a0P (B) + P (A) \u2013\u00a0P (A\u00a0\u222a\u00a0B)<\/p>\n<p>By using data from table, we get:<\/p>\n<p>\u2234\u00a0P (A\u00a0\u2229\u00a0B) = 0.5 + 0.35 \u2013 0.7<\/p>\n<p>= 0.15<\/p>\n<p>Hence the table is:<\/p>\n<p><img class=\"alignnone size-full wp-image-121519\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/d.png\" alt=\"d\" width=\"341\" height=\"110\" \/><\/p>\n<p><strong>2. If A and B are two events associated with a random experiment such that P (A) = 0.3, P (B) = 0.4 and P (A\u00a0\u222a\u00a0B) = 0.5, find P (A\u00a0\u2229\u00a0B).<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given: A and B are two events.<\/p>\n<p>P (A) = 0.3, P (B) = 0.5 and P (A\u00a0\u222a\u00a0B) = 0.5<\/p>\n<p>Now we need to find P (A\u00a0\u2229\u00a0B).<\/p>\n<p>By definition of P (A or B) under axiomatic approach (also called addition theorem) we know that:<\/p>\n<p>P (A\u00a0\u222a\u00a0B) = P (A) + P (B) \u2013 P (A\u00a0\u2229\u00a0B)<\/p>\n<p>So,\u00a0P (A\u00a0\u2229\u00a0B) = P (A) + P (B) \u2013 P (A\u00a0\u222a\u00a0B)<\/p>\n<p>P (A\u00a0\u2229\u00a0B) = 0.3 + 0.4 \u2013 0.5<\/p>\n<p>= 0.7 \u2013 0.5<\/p>\n<p>= 0.2<\/p>\n<p>\u2234\u00a0P (A\u00a0\u2229\u00a0B) is 0.2<\/p>\n<p><strong>3. If A and B are two events associated with a random experiment such that P (A) = 0.5, P (B) = 0.3 and P (A\u00a0\u2229\u00a0B) = 0.2, find P (A\u00a0\u222a\u00a0B).<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given: A and B are two events.<\/p>\n<p>P (A) = 0.5, P (B) = 0.3 and P (A\u00a0\u2229\u00a0B) = 0.2<\/p>\n<p>Now we need to find P (A\u00a0\u222a\u00a0B).<\/p>\n<p>By definition of P (A or B) under axiomatic approach (also called addition theorem) we know that:<\/p>\n<p>P (A\u00a0\u222a\u00a0B) = P (A) + P (B) \u2013 P (A\u00a0\u2229\u00a0B)<\/p>\n<p>So, P (A\u00a0\u222a\u00a0B) = P (A) + P (B) \u2013 P (A\u00a0\u2229\u00a0B)<\/p>\n<p>P (A\u00a0\u222a\u00a0B) = 0.5 + 0.3 \u2013 0.2<\/p>\n<p>= 0.8 \u2013 0.2<\/p>\n<p>= 0.6<\/p>\n<p>\u2234\u00a0P (A\u00a0\u222a\u00a0B) is 0.6<\/p>\n<p><strong>Probability Ex 33.4 Q5<\/strong><br \/><img src=\"https:\/\/farm9.staticflickr.com\/8612\/16112166042_280ff8917c_o.png\" alt=\"RD-Sharma-class-11 Solutions-Chapter-33-Probability-Ex-33.4-Q-5\" width=\"491\" height=\"265\" \/><br \/><strong>Probability Ex 33.4 Q6<\/strong><br \/><img src=\"https:\/\/farm8.staticflickr.com\/7556\/15927144577_341331341f_o.png\" alt=\"RD-Sharma-class-11 Solutions-Chapter-33-Probability-Ex-33.4-Q-6\" width=\"437\" height=\"626\" \/><br \/><strong>Probability Ex 33.4 Q7<\/strong><br \/><img src=\"https:\/\/farm8.staticflickr.com\/7497\/15493203813_cd61bd7cf7_o.png\" alt=\"RD-Sharma-class-11 Solutions-Chapter-33-Probability-Ex-33.4-Q-7\" width=\"374\" height=\"195\" \/><br \/><strong>Probability Ex 33.4 Q8<\/strong><br \/><img src=\"https:\/\/farm8.staticflickr.com\/7553\/16110986091_8f085c7336_o.png\" alt=\"RD-Sharma-class-11 Solutions-Chapter-33-Probability-Ex-33.4-Q-8\" width=\"536\" height=\"376\" \/><br \/><strong>Probability Ex 33.4 Q9<\/strong><br \/><img src=\"https:\/\/farm9.staticflickr.com\/8582\/15490577614_d7c97a4a56_o.png\" alt=\"RD-Sharma-class-11 Solutions-Chapter-33-Probability-Ex-33.4-Q-9\" width=\"420\" height=\"495\" \/><br \/><strong>Probability Ex 33.4 Q10<\/strong><br \/><img src=\"https:\/\/farm8.staticflickr.com\/7521\/16087110266_d4469f5f0a_o.png\" alt=\"RD-Sharma-class-11 Solutions-Chapter-33-Probability-Ex-33.4-Q-10\" width=\"527\" height=\"440\" \/><br \/><strong>Probability Ex 33.4 Q11<\/strong><br \/><img src=\"https:\/\/farm9.staticflickr.com\/8644\/15927144097_fc50f29d35_o.png\" alt=\"RD-Sharma-class-11 Solutions-Chapter-33-Probability-Ex-33.4-Q-11\" width=\"596\" height=\"299\" \/><br \/><strong>Probability Ex 33.4 Q12<\/strong><br \/><img src=\"https:\/\/farm8.staticflickr.com\/7466\/16112165242_1c0610dd69_o.png\" alt=\"RD-Sharma-class-11 Solutions-Chapter-33-Probability-Ex-33.4-Q-12\" width=\"549\" height=\"282\" \/><br \/><strong>Probability Ex 33.4 Q13<\/strong><br \/><img src=\"https:\/\/farm8.staticflickr.com\/7571\/15926832569_70d5672d5a_o.png\" alt=\"RD-Sharma-class-11 Solutions-Chapter-33-Probability-Ex-33.4-Q-13\" width=\"426\" height=\"302\" \/><br \/><strong>Probability Ex 33.4 Q14<\/strong><br \/><img src=\"https:\/\/farm8.staticflickr.com\/7554\/15925477268_a03381d0b7_o.png\" alt=\"RD-Sharma-class-11 Solutions-Chapter-33-Probability-Ex-33.4-Q-14\" width=\"416\" height=\"384\" \/><br \/><strong>Probability Ex 33.4 Q15<\/strong><br \/><img src=\"https:\/\/farm8.staticflickr.com\/7467\/15490576874_a778f55036_o.png\" alt=\"RD-Sharma-class-11 Solutions-Chapter-33-Probability-Ex-33.4-Q-15\" width=\"436\" height=\"236\" \/><br \/><strong>Probability Ex 33.4 Q16<\/strong><br \/><img src=\"https:\/\/farm9.staticflickr.com\/8616\/16112164752_f96b9b2b2e_o.png\" alt=\"RD-Sharma-class-11 Solutions-Chapter-33-Probability-Ex-33.4-Q-16\" width=\"589\" height=\"313\" \/><br \/><strong>Probability Ex 33.4 Q17<\/strong><br \/><img src=\"https:\/\/farm9.staticflickr.com\/8597\/16087109516_873914b50f_o.png\" alt=\"RD-Sharma-class-11 Solutions-Chapter-33-Probability-Ex-33.4-Q-17\" width=\"633\" height=\"289\" \/><br \/><strong>Probability Ex 33.4 Q18<\/strong><br \/><img src=\"https:\/\/farm8.staticflickr.com\/7549\/16112164592_8b994638a1_o.png\" alt=\"RD-Sharma-class-11 Solutions-Chapter-33-Probability-Ex-33.4-Q-18\" width=\"492\" height=\"464\" \/><br \/><strong>Probability Ex 33.4 Q19<\/strong><br \/><img src=\"https:\/\/farm8.staticflickr.com\/7498\/16110984811_61a66ed683_o.png\" alt=\"RD-Sharma-class-11 Solutions-Chapter-33-Probability-Ex-33.4-Q-19\" width=\"372\" height=\"240\" \/><br \/><strong>Probability Ex 33.4 Q20<\/strong><br \/><img src=\"https:\/\/farm8.staticflickr.com\/7492\/15493202413_35fd05725a_o.png\" alt=\"RD-Sharma-class-11 Solutions-Chapter-33-Probability-Ex-33.4-Q-20\" width=\"581\" height=\"505\" \/><br \/><strong>Probability Ex 33.4 Q21<\/strong><br \/><img src=\"https:\/\/farm8.staticflickr.com\/7547\/16112878825_d55c5d7d61_o.png\" alt=\"RD-Sharma-class-11 Solutions-Chapter-33-Probability-Ex-33.4-Q-21\" width=\"473\" height=\"307\" \/><br \/><strong>Probability Ex 33.4 Q22<\/strong><br \/><img src=\"https:\/\/farm8.staticflickr.com\/7568\/15493202343_b15c0cd782_o.png\" alt=\"RD-Sharma-class-11 Solutions-Chapter-33-Probability-Ex-33.4-Q-22\" width=\"484\" height=\"580\" \/><br \/><strong>Probability Ex 33.4 Q23<\/strong><br \/><img src=\"https:\/\/farm9.staticflickr.com\/8601\/15493202333_d8cc1336a5_o.png\" alt=\"RD-Sharma-class-11 Solutions-Chapter-33-Probability-Ex-33.4-Q-23\" width=\"428\" height=\"366\" \/><br \/><strong>Probability Ex 33.4 Q24<\/strong><br \/><img src=\"https:\/\/farm8.staticflickr.com\/7560\/16112164372_3c423e8a83_o.png\" alt=\"RD-Sharma-class-11 Solutions-Chapter-33-Probability-Ex-33.4-Q-24\" width=\"375\" height=\"360\" \/><br \/><strong>Probability Ex 33.4 Q25<\/strong><br \/><img src=\"https:\/\/farm8.staticflickr.com\/7511\/15490576374_fc17b7b7f0_o.png\" alt=\"RD-Sharma-class-11 Solutions-Chapter-33-Probability-Ex-33.4-Q-25\" width=\"408\" height=\"176\" \/><br \/><strong>Probability Ex 33.4 Q26<\/strong><br \/><img src=\"https:\/\/farm8.staticflickr.com\/7540\/15925597590_fea135831e_o.png\" alt=\"RD-Sharma-class-11 Solutions-Chapter-33-Probability-Ex-33.4-Q-26\" width=\"231\" height=\"69\" \/><br \/><strong>Probability Ex 33.4 Q27<\/strong><br \/><img src=\"https:\/\/farm9.staticflickr.com\/8670\/16112164272_7c75830592_o.png\" alt=\"RD-Sharma-class-11 Solutions-Chapter-33-Probability-Ex-33.4-Q-27\" width=\"337\" height=\"126\" \/><br \/><strong>Probability Ex 33.4 Q28<\/strong><br \/><img src=\"https:\/\/farm9.staticflickr.com\/8663\/15493202263_b2b1667640_o.png\" alt=\"RD-Sharma-class-11 Solutions-Chapter-33-Probability-Ex-33.4-Q-28\" width=\"352\" height=\"291\" \/><\/p>\n<p>We have provided complete details of RD Sharma Solutions Class 11 Maths Chapter 33 Exercise 33.4. If you have any queries related to <a href=\"https:\/\/www.cbse.gov.in\/\" target=\"_blank\" rel=\"noopener\"><strong>CBSE<\/strong><\/a>\u00a0Class 11, feel free to ask us in the comment section below.<\/p>\n<h2><span class=\"ez-toc-section\" id=\"faqs-on-rd-sharma-solutions-class-11-maths-chapter-33-exercise-334\"><\/span>FAQs on RD Sharma Solutions Class 11 Maths Chapter 33 Exercise 33.4<span class=\"ez-toc-section-end\"><\/span><\/h2>\n\n\n<div id=\"rank-math-faq\" class=\"rank-math-block\">\n<div class=\"rank-math-list \">\n<div id=\"faq-question-1630320329518\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"how-many-questions-are-there-in-rd-sharma-class-11-solutions-chapter-33-statistics-exercise-334\"><\/span>How many questions are there in RD Sharma Class 11 Solutions Chapter 33 Statistics Exercise 33.4?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>There are a total of 9 questions in RD Sharma Solutions Class 11 Maths Chapter 33 Exercise 33.4.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1630320362871\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"what-are-the-advantages-of-using-rd-sharma-solutions-class-11-maths-chapter-33-exercise-334\"><\/span>What are the advantages of using RD Sharma Solutions Class 11 Maths Chapter 33 Exercise 33.4?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>Referring to RD Sharma Solutions Class 11 Maths Chapter 33 Exercise 33.4 will provide students a good understanding of the type of questions that might be asked in the Class 11 exam from the chapter.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1630320519913\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"where-can-i-download-rd-sharma-solutions-class-11-maths-chapter-33-exercise-334-free-pdf\"><\/span>Where can I download RD Sharma Solutions Class 11 Maths Chapter 33 Exercise 33.4 free PDF?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>You can download RD Sharma Class 11 Solutions Chapter 33 Exercise 33.4 free PDF from the above article.<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<\/div>","protected":false},"excerpt":{"rendered":"<p>RD Sharma Solutions Class 11 Maths Chapter 33 Exercise 33.4:\u00a0Until now, we&#8217;ve computed the likelihood of an event occurring or not occurring using the favourable and total number of elementary events. The RD Sharma Class 11 Solutions are described in simple language by topic experts to assist students in improving their analytical thinking and problem-solving &#8230; <a title=\"RD Sharma Class 11 Solutions Chapter 33 Probability Exercise 33.4 (Updated for 2021-22)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-33-exercise-33-4\/\" aria-label=\"More on RD Sharma Class 11 Solutions Chapter 33 Probability Exercise 33.4 (Updated for 2021-22)\">Read more<\/a><\/p>\n","protected":false},"author":238,"featured_media":121471,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"","fifu_image_alt":""},"categories":[73411,2985,73410],"tags":[3428,73334,73564],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/68139"}],"collection":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/users\/238"}],"replies":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/comments?post=68139"}],"version-history":[{"count":5,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/68139\/revisions"}],"predecessor-version":[{"id":121536,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/68139\/revisions\/121536"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/121471"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=68139"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=68139"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=68139"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}