{"id":33664,"date":"2019-08-19T13:13:06","date_gmt":"2019-08-19T07:43:06","guid":{"rendered":"https:\/\/cracku.in\/blog\/?p=33664"},"modified":"2019-08-19T13:13:06","modified_gmt":"2019-08-19T07:43:06","slug":"probability-questions-for-rrb-ntpc-pdf","status":"publish","type":"post","link":"https:\/\/cracku.in\/blog\/probability-questions-for-rrb-ntpc-pdf\/","title":{"rendered":"Probability Questions for RRB NTPC PDF"},"content":{"rendered":"<h1><span style=\"text-decoration: underline;\"><strong>Probability Questions for RRB NTPC PDF<\/strong><\/span><\/h1>\n<p>Download RRB NTPC Probability Questions and Answers PDF. Top 15 RRB NTPC Probability questions based on asked questions in previous exam papers very important for the Railway NTPC exam.<\/p>\n<p class=\"text-center\"><a href=\"https:\/\/cracku.in\/downloads\/5968\" target=\"_blank\" class=\"btn btn-danger  download\">Download Probability Questions for RRB NTPC PDF<\/a><\/p>\n<!-- Error, Advert is not available at this time due to schedule\/geolocation restrictions! -->\n<p>Take a <a href=\"https:\/\/cracku.in\/rrb-ntpc-mock-test\" target=\"_blank\" rel=\"noopener\">free mock test for RRB NTPC<\/a><\/p>\n<p>Download <a href=\"https:\/\/cracku.in\/railways-ntpc-previous-papers\" target=\"_blank\" rel=\"noopener\">RRB NTPC Previous Papers PDF<\/a><\/p>\n<p><b>Question 1:\u00a0<\/b>The Indian cricket team is to be selected out of fifteen players five of them are bowlers In how many ways the team can be selected so that the team contains at least three bowlers ?<\/p>\n<p>a)\u00a01260<\/p>\n<p>b)\u00a01620<\/p>\n<p>c)\u00a01250<\/p>\n<p>d)\u00a01200<\/p>\n<p><b>Question 2:\u00a0<\/b>How many different words can be made from the five vowels such that exactly one letter is repeated?<\/p>\n<p>a)\u00a0360<\/p>\n<p>b)\u00a0720<\/p>\n<p>c)\u00a01440<\/p>\n<p>d)\u00a01800<\/p>\n<p><b>Question 3:\u00a0<\/b>If three coins are tossed, what is the probability that atleast 2 heads occur?<\/p>\n<p>a)\u00a00.5<\/p>\n<p>b)\u00a00.25<\/p>\n<p>c)\u00a00.2<\/p>\n<p>d)\u00a00.1<\/p>\n<p class=\"text-center\"><a href=\"https:\/\/cracku.in\/railways-ntpc-previous-papers\" target=\"_blank\" class=\"btn btn-primary \">RRB NTPC Previous Papers [Download PDF]<\/a><\/p>\n<p class=\"text-center\"><a href=\"https:\/\/www.youtube.com\/channel\/UCMDJPaiDdRPv2mrEJoLfklA?sub_confirmation=1\" target=\"_blank\" class=\"btn btn-warning \">FREE RRB NTPC YOUTUBE VIDEOS<\/a><\/p>\n<p><b>Question 4:\u00a0<\/b>Find the probability that at least 3 heads appear when 4 unbiased coins are tossed.<\/p>\n<p>a)\u00a0$\\frac{11}{16}$<\/p>\n<p>b)\u00a0$\\frac{5}{16}$<\/p>\n<p>c)\u00a0$\\frac{3}{8}$<\/p>\n<p>d)\u00a0$\\frac{5}{8}$<\/p>\n<p><b>Question 5:\u00a0<\/b>Two dice are thrown and the numbers that appear on them are a and b respectively. Find the probability that the sum of a and b is 8.<\/p>\n<p>a)\u00a0$\\frac{5}{6}$<\/p>\n<p>b)\u00a0$\\frac{1}{6}$<\/p>\n<p>c)\u00a0$\\frac{5}{36}$<\/p>\n<p>d)\u00a0$\\frac{31}{36}$<\/p>\n<p><b>Question 6:\u00a0<\/b>If I pick up a card from a well-shuffled pack of cards, What will be the probability of getting an ace?<\/p>\n<p>a)\u00a01\/20<\/p>\n<p>b)\u00a01\/15<\/p>\n<p>c)\u00a01\/13<\/p>\n<p>d)\u00a01\/16<\/p>\n<p><b>Question 7:\u00a0<\/b>What is the probability of getting 3 face cards if three cards are drawn from a set of 52 playing cards?<\/p>\n<p>a)\u00a0$\\frac{11}{1105}$<\/p>\n<p>b)\u00a0$\\frac{7}{1105}$<\/p>\n<p>c)\u00a0$\\frac{17}{1105}$<\/p>\n<p>d)\u00a0$\\frac{19}{1105}$<\/p>\n<p class=\"text-center\"><a href=\"https:\/\/cracku.in\/rrb-ntpc-mock-test\" target=\"_blank\" class=\"btn btn-danger \">RRB NTPC Free Mock Test<\/a><\/p>\n<!-- Error, Advert is not available at this time due to schedule\/geolocation restrictions! -->\n<p><b>Question 8:\u00a0<\/b>In how many ways can we arrange the letters of the COMMIT so that the vowels are at the extreme ends?<\/p>\n<p>a)\u00a014<\/p>\n<p>b)\u00a024<\/p>\n<p>c)\u00a06!\/2!<\/p>\n<p>d)\u00a036<\/p>\n<p><b>Question 9:\u00a0<\/b>In a wooden box, there are four red and four black balls. If two balls are taken out simultaneously at random, then the probability of both balls being red is:<\/p>\n<p>a)\u00a0$ \\frac{1}{2} $<\/p>\n<p>b)\u00a0$ \\frac{1}{4} $<\/p>\n<p>c)\u00a0$ \\frac{3}{8} $<\/p>\n<p>d)\u00a0$ \\frac{3}{14} $<\/p>\n<p><b>Question 10:\u00a0<\/b>A complete cycle of a traffic lights takes 60 seconds. During each cycle the light is green for 25 seconds, yellow for 5 seconds and red for 30 seconds. At a randomly chosen time, the probability that the light will not be green is:<\/p>\n<p>a)\u00a0$ \\frac{1}{3} $<\/p>\n<p>b)\u00a0$ \\frac{1}{4} $<\/p>\n<p>c)\u00a0$ \\frac{5}{12} $<\/p>\n<p>d)\u00a0$ \\frac{7}{12} $<\/p>\n<p><b>Question 11:\u00a0<\/b>A fair die is rolled twice. What is the probability that the sum of the two outcomes is more than 6?<\/p>\n<p>a)\u00a0$\\frac{2}{9}$<\/p>\n<p>b)\u00a0$\\frac{1}{4}$<\/p>\n<p>c)\u00a0$\\frac{7}{12}$<\/p>\n<p>d)\u00a0$\\frac{4}{9}$<\/p>\n<p class=\"text-center\"><a href=\"https:\/\/cracku.in\/blog\/general-science-questions-answers-competitive-exams-pdf-mcq-quiz\/\" target=\"_blank\" class=\"btn btn-info \">Download General Science Notes PDF<\/a><\/p>\n<!-- Error, Advert is not available at this time due to schedule\/geolocation restrictions! -->\n<p><b>Question 12:\u00a0<\/b>Let P (n) be the probability that the outcome is \u2018n\u2019 when a die is rolled. A biased die is such that,<br \/>\n&#8211; P (1) = P (3) = P (5) = x<br \/>\n&#8211; P (2) = P (4) = P (6) = y.<br \/>\n&#8211; y = 2x.<br \/>\nWhat is the probability that the outcome is a prime when the die is rolled?<\/p>\n<p>a)\u00a0$\\frac{4}{9}$<\/p>\n<p>b)\u00a0$\\frac{5}{9}$<\/p>\n<p>c)\u00a0$\\frac{1}{2}$<\/p>\n<p>d)\u00a0$\\frac{1}{3}$<\/p>\n<p><b>Question 13:\u00a0<\/b>Two fair dice are rolled simultaneously. What is the probability that atleast one of the dice shows up a three?<\/p>\n<p>a)\u00a0$\\frac{5}{6}$<\/p>\n<p>b)\u00a0$\\frac{11}{36}$<\/p>\n<p>c)\u00a0$\\frac{1}{2}$<\/p>\n<p>d)\u00a0$\\frac{25}{36}$<\/p>\n<p><b>Question 14:\u00a0<\/b>Rajat and Piyush were playing Scrabbles. Piyush made the word \u2018BUMFUZZLED\u2019. Rajat took 5 letters from this at random and arranged them randomly. What is the probability that he made the word \u2018FUMED\u2019?<\/p>\n<p>a)\u00a0$\\frac{1}{126}$<\/p>\n<p>b)\u00a0$\\frac{1}{3024}$<\/p>\n<p>c)\u00a0$\\frac{5}{9072}$<\/p>\n<p>d)\u00a0None of these<\/p>\n<p><b>Question 15:\u00a0<\/b>A real number $k$ is chosen from 0 to 15. What is the probability that $k^2$ $\\geq$49?<\/p>\n<p>a)\u00a08\/15<\/p>\n<p>b)\u00a01\/2<\/p>\n<p>c)\u00a07\/16<\/p>\n<p>d)\u00a0none of these<\/p>\n<p class=\"text-center\"><a href=\"https:\/\/cracku.in\/rrb-ntpc-mock-test\" target=\"_blank\" class=\"btn btn-primary \">RRB NTPC Free Mock Test<\/a><\/p>\n<p class=\"text-center\"><a href=\"https:\/\/cracku.in\/railways-ntpc-previous-papers\" target=\"_blank\" class=\"btn btn-danger \">RRB NTPC Previous Papers (Download PDF)<\/a><\/p>\n<!-- Error, Advert is not available at this time due to schedule\/geolocation restrictions! -->\n<p><span style=\"text-decoration: underline;\"><strong>Answers &amp; Solutions:<\/strong><\/span><\/p>\n<p><strong>1)\u00a0Answer\u00a0(A)<\/strong><\/p>\n<p>To select a team of 11, there can be 5 bowlers and 6 batsmen or 4 bowlers and 7 batsmen or 3 bowlers and 8 batsmen. This can be done in $C_6^{10} + C_4^5 \\times C_7^{10} + C_3^5 \\times C_8^{10}$<\/p>\n<p>= 1260<\/p>\n<p><strong>2)\u00a0Answer\u00a0(D)<\/strong><\/p>\n<p>In total, there are 6 letters of which one is repeated.<br \/>\nNumber of ways of arranging 6 letters is 6! = 720<\/p>\n<p>One vowel out of 5 can be selected in 5 ways.<br \/>\nOne repitition =&gt; 5*720\/2! = 1800<\/p>\n<p><strong>3)\u00a0Answer\u00a0(A)<\/strong><\/p>\n<p>The total number of possibilities are TTT, TTH, THT, HTT, THH, HTH, HHT and HHH.<br \/>\nWe need to select those cases in which there are at least 2 Hs.<br \/>\nThere are 4 such cases out of 8.<br \/>\n=&gt; Probability = 4\/8 = 0.5<\/p>\n<p><strong>4)\u00a0Answer\u00a0(B)<\/strong><\/p>\n<p>Favourable cases are HHHT, HHTH, HTHH, THHH and HHHH =&gt; 5<br \/>\nTotal number of cases are 2*2*2*2 = 16<br \/>\nProbability = $\\frac{5}{16}$<\/p>\n<p><strong>5)\u00a0Answer\u00a0(C)<\/strong><\/p>\n<p>Number of ways such that sum is 8 =&gt; (2,6), (3,5), (4,4), (5,3), (6,2)<br \/>\n=&gt; 5 possible cases.<br \/>\nTotal number of cases = 6*6 = 36<br \/>\nProbability = $\\frac{5}{36}$<\/p>\n<p><strong>6)\u00a0Answer\u00a0(C)<\/strong><\/p>\n<p>As there are a total of 4 ace cards out of 52 cards hence probability will be = 4\/52 = 1\/13<\/p>\n<p><strong>7)\u00a0Answer\u00a0(A)<\/strong><\/p>\n<p>There are 12 face cards in each deck : 4 Kings, 4 Queens and 4 Jacks. We can select 3 out of these in $^{12} C _3$ ways. We can select 3 cards in $^{52} C _3$ ways. Hence, probability = $^{12} C _3$ \/ $^{52} C _3$ = 12*11*10\/52*51*50 = 11\/1105.<\/p>\n<p><strong>8)\u00a0Answer\u00a0(B)<\/strong><\/p>\n<p>The vowels would be at either ends. The remaining letters can be arranged in 4!\/2! = 12 ways. The vowels can be arranged among themselves in 2! =2 ways. Total number of ways = 12*2=24 ways.<\/p>\n<p><strong>9)\u00a0Answer\u00a0(D)<\/strong><\/p>\n<p>Probability of both balls being red=$\\frac{4}{8}$\u00d7$\\frac{3}{7}$=3\/14<\/p>\n<p><strong>10)\u00a0Answer\u00a0(D)<\/strong><\/p>\n<p>probability that the light is not green=probablity that the light is green or red=(30 +5)\/60=7\/12<\/p>\n<p><strong>11)\u00a0Answer\u00a0(C)<\/strong><\/p>\n<p>The favourable cases are when the sum is 7, 8, 9, 10, 11 and 12.<br \/>\nWe can use the 6&#215;6 matrix to find out the number of ways in which the sum can be 7, 8, 9, 10, 11 and 12.<br \/>\nThese are as highlighted in green.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"img-responsive\" src=\"https:\/\/cracku.in\/media\/uploads\/10.PNG\" width=\"300\" height=\"261\" \/><\/p>\n<p>Thus the number of ways = 6+5+4+3+2+1 = 21.<br \/>\nProbability = 21\/36 = 7\/12.<\/p>\n<p><strong>12)\u00a0Answer\u00a0(A)<\/strong><\/p>\n<p>We know that,<br \/>\nP(1)+P(2)+P(3)+P(4)+P(5)+P(6) = 1<br \/>\n=&gt; 3x+3y = 1<br \/>\n=&gt; 3x+6x = 1<br \/>\n=&gt; x = 1\/9<br \/>\nThus,<br \/>\nP (1) = P (3) = P (5) = 1\/9<br \/>\nP (2) = P (4) = P (6) = 2\/9<br \/>\nThe favourable outcomes are 2, 3 and 5.<br \/>\nRequired Probability = P(2) +P(3) +P(5) = 4\/9.<\/p>\n<p><strong>13)\u00a0Answer\u00a0(B)<\/strong><\/p>\n<p>We can use the 6&#215;6 matrix to find out the number of ways in which atleast one of the dice shows up a three.<br \/>\nThe favourable cells are as highlighted in green.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"img-responsive\" src=\"https:\/\/cracku.in\/media\/uploads\/9.PNG\" width=\"304\" height=\"260\" \/><\/p>\n<p>Thus the number of favourable ways = 11.<br \/>\nProbability = 11\/36.<\/p>\n<p><strong>14)\u00a0Answer\u00a0(D)<\/strong><\/p>\n<p>The probability that he took the letters- F, U, M, E, and D is $\\frac{2}{^{10}C_{5}}=\\frac{2}{252}=\\frac{1}{126}$. [2 Us are available]<br \/>\nAfter selecting the letters, the probability that he arranged them in this particular order is $\\frac{1}{5!}$<br \/>\nSo, the probability that he made \u2018FUMED\u2019 is $\\frac{1}{126} \\times \\frac{1}{5!} = \\frac{1}{15120}$<\/p>\n<p><strong>15)\u00a0Answer\u00a0(A)<\/strong><\/p>\n<p>We can divide the numbers from 0 to 15 into 15 equal ranges i.e. 0-1, 1-2&#8230;.till 14-15.<br \/>\n$k^2 \\geq$ 7 only from 7 to 15. From 7 &#8211; 15, there are 8 equal ranges.<br \/>\nThus, the required probability = favorable outcomes\/total outcomes = 8\/15<\/p>\n<!-- Error, Advert is not available at this time due to schedule\/geolocation restrictions! -->\n<p class=\"text-center\"><a href=\"https:\/\/play.google.com\/store\/apps\/details?id=in.cracku.app&amp;hl=en_US\" target=\"_blank\" class=\"btn btn-danger \">DOWNLOAD APP FOR RRB FREE MOCKS<\/a><\/p>\n<p>We hope this Probability Questions pdf\u00a0 for RRB NTPC Exam will be highly useful for your Preparation.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Probability Questions for RRB NTPC PDF Download RRB NTPC Probability Questions and Answers PDF. Top 15 RRB NTPC Probability questions based on asked questions in previous exam papers very important for the Railway NTPC exam. Take a free mock test for RRB NTPC Download RRB NTPC Previous Papers PDF Question 1:\u00a0The Indian cricket team is [&hellip;]<\/p>\n","protected":false},"author":41,"featured_media":33669,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"om_disable_all_campaigns":false,"_mi_skip_tracking":false,"footnotes":""},"categories":[169,125,31,1603,1601],"tags":[491,1637,1647,1619],"class_list":{"0":"post-33664","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-downloads","8":"category-featured","9":"category-railways","10":"category-rrb-ntpc","11":"category-rrb-ntpc-je","12":"tag-rrb","13":"tag-rrb-exam","14":"tag-rrb-mocks","15":"tag-rrb-ntpc-2019"},"better_featured_image":{"id":33669,"alt_text":"Probability Questions for RRB NTPC PDF","caption":"Probability Questions for RRB NTPC PDF","description":"Probability Questions for RRB NTPC 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Top 15 RRB NTPC Probability questions based on asked questions in previous exam papers very important for the Railway NTPC exam. 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