{"id":93132,"date":"2025-06-01T11:42:30","date_gmt":"2025-06-01T11:42:30","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=93132"},"modified":"2025-06-01T11:42:30","modified_gmt":"2025-06-01T11:42:30","slug":"from-a-pack-of-5-green-balls-and-4-red-balls-2-balls-are-drawn-at-ran","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/from-a-pack-of-5-green-balls-and-4-red-balls-2-balls-are-drawn-at-ran\/","title":{"rendered":"From a pack of 5 green balls and 4 red balls, 2 balls are drawn at ran"},"content":{"rendered":"

From a pack of 5 green balls and 4 red balls, 2 balls are drawn at random. What is the probability that both the balls are of the same colour ?<\/p>\n

[amp_mcq option1=”5\/9″ option2=”4\/9″ option3=”2\/9″ option4=”None of the above” correct=”option2″]<\/p>\n

\n
\n
This question was previously asked in<\/div>\n
UPSC CISF-AC-EXE – 2022<\/div>\n<\/div>\n
Download PDF<\/a>Attempt Online<\/a><\/div>\n<\/div>\n
\nThe correct answer is 4\/9.
\n<\/section>\n
\nTotal number of balls in the pack is 5 green + 4 red = 9 balls.
\nWe are drawing 2 balls at random.
\nThe total number of ways to choose 2 balls out of 9 is given by the combination formula C(n, k) = n! \/ (k! * (n-k)!):
\nTotal outcomes = C(9, 2) = 9! \/ (2! * 7!) = (9 \u00d7 8) \/ (2 \u00d7 1) = 36.
\nWe want the probability that both balls are of the same colour. This can happen in two mutually exclusive ways:
\nCase 1: Both balls are green.
\nNumber of ways to choose 2 green balls out of 5 = C(5, 2) = 5! \/ (2! * 3!) = (5 \u00d7 4) \/ (2 \u00d7 1) = 10.
\nCase 2: Both balls are red.
\nNumber of ways to choose 2 red balls out of 4 = C(4, 2) = 4! \/ (2! * 2!) = (4 \u00d7 3) \/ (2 \u00d7 1) = 6.
\nThe number of favourable outcomes (both balls of the same colour) = Number of ways (both green) + Number of ways (both red) = 10 + 6 = 16.
\nThe probability that both balls are of the same colour is (Favourable outcomes) \/ (Total outcomes) = 16 \/ 36.
\nSimplifying the fraction, 16\/36 = (4 \u00d7 4) \/ (9 \u00d7 4) = 4\/9.
\n<\/section>\n
\nThis problem involves basic concepts of probability and combinations. The calculation for combinations C(n, k) represents the number of ways to choose k items from a set of n distinct items without regard to the order of selection. When dealing with probabilities of multiple events, we sum the probabilities of mutually exclusive events (like drawing two green OR two red).
\n<\/section>\n","protected":false},"excerpt":{"rendered":"

From a pack of 5 green balls and 4 red balls, 2 balls are drawn at random. What is the probability that both the balls are of the same colour ? [amp_mcq option1=”5\/9″ option2=”4\/9″ option3=”2\/9″ option4=”None of the above” correct=”option2″] This question was previously asked in UPSC CISF-AC-EXE – 2022 Download PDFAttempt Online The correct … <\/p>\n

Detailed SolutionFrom a pack of 5 green balls and 4 red balls, 2 balls are drawn at ran<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1089],"tags":[1108,1102],"class_list":["post-93132","post","type-post","status-publish","format-standard","hentry","category-upsc-cisf-ac-exe","tag-1108","tag-quantitative-aptitude-and-reasoning","no-featured-image-padding"],"yoast_head":"\nFrom a pack of 5 green balls and 4 red balls, 2 balls are drawn at ran<\/title>\n<meta name=\"description\" content=\"The correct answer is 4\/9. Total number of balls in the pack is 5 green + 4 red = 9 balls. We are drawing 2 balls at random. The total number of ways to choose 2 balls out of 9 is given by the combination formula C(n, k) = n! \/ (k! * (n-k)!): Total outcomes = C(9, 2) = 9! \/ (2! * 7!) = (9 \u00d7 8) \/ (2 \u00d7 1) = 36. We want the probability that both balls are of the same colour. This can happen in two mutually exclusive ways: Case 1: Both balls are green. Number of ways to choose 2 green balls out of 5 = C(5, 2) = 5! \/ (2! * 3!) = (5 \u00d7 4) \/ (2 \u00d7 1) = 10. Case 2: Both balls are red. Number of ways to choose 2 red balls out of 4 = C(4, 2) = 4! \/ (2! * 2!) = (4 \u00d7 3) \/ (2 \u00d7 1) = 6. The number of favourable outcomes (both balls of the same colour) = Number of ways (both green) + Number of ways (both red) = 10 + 6 = 16. The probability that both balls are of the same colour is (Favourable outcomes) \/ (Total outcomes) = 16 \/ 36. Simplifying the fraction, 16\/36 = (4 \u00d7 4) \/ (9 \u00d7 4) = 4\/9.\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/exam.pscnotes.com\/mcq\/from-a-pack-of-5-green-balls-and-4-red-balls-2-balls-are-drawn-at-ran\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"From a pack of 5 green balls and 4 red balls, 2 balls are drawn at ran\" \/>\n<meta property=\"og:description\" content=\"The correct answer is 4\/9. Total number of balls in the pack is 5 green + 4 red = 9 balls. We are drawing 2 balls at random. The total number of ways to choose 2 balls out of 9 is given by the combination formula C(n, k) = n! \/ (k! * (n-k)!): Total outcomes = C(9, 2) = 9! \/ (2! * 7!) = (9 \u00d7 8) \/ (2 \u00d7 1) = 36. We want the probability that both balls are of the same colour. This can happen in two mutually exclusive ways: Case 1: Both balls are green. Number of ways to choose 2 green balls out of 5 = C(5, 2) = 5! \/ (2! * 3!) = (5 \u00d7 4) \/ (2 \u00d7 1) = 10. Case 2: Both balls are red. Number of ways to choose 2 red balls out of 4 = C(4, 2) = 4! \/ (2! * 2!) = (4 \u00d7 3) \/ (2 \u00d7 1) = 6. The number of favourable outcomes (both balls of the same colour) = Number of ways (both green) + Number of ways (both red) = 10 + 6 = 16. The probability that both balls are of the same colour is (Favourable outcomes) \/ (Total outcomes) = 16 \/ 36. Simplifying the fraction, 16\/36 = (4 \u00d7 4) \/ (9 \u00d7 4) = 4\/9.\" \/>\n<meta property=\"og:url\" content=\"https:\/\/exam.pscnotes.com\/mcq\/from-a-pack-of-5-green-balls-and-4-red-balls-2-balls-are-drawn-at-ran\/\" \/>\n<meta property=\"og:site_name\" content=\"MCQ and Quiz for Exams\" \/>\n<meta property=\"article:published_time\" content=\"2025-06-01T11:42:30+00:00\" \/>\n<meta name=\"author\" content=\"rawan239\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"rawan239\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"1 minute\" \/>\n<!-- \/ Yoast SEO Premium plugin. -->","yoast_head_json":{"title":"From a pack of 5 green balls and 4 red balls, 2 balls are drawn at ran","description":"The correct answer is 4\/9. Total number of balls in the pack is 5 green + 4 red = 9 balls. We are drawing 2 balls at random. The total number of ways to choose 2 balls out of 9 is given by the combination formula C(n, k) = n! \/ (k! * (n-k)!): Total outcomes = C(9, 2) = 9! \/ (2! * 7!) = (9 \u00d7 8) \/ (2 \u00d7 1) = 36. We want the probability that both balls are of the same colour. This can happen in two mutually exclusive ways: Case 1: Both balls are green. Number of ways to choose 2 green balls out of 5 = C(5, 2) = 5! \/ (2! * 3!) = (5 \u00d7 4) \/ (2 \u00d7 1) = 10. Case 2: Both balls are red. Number of ways to choose 2 red balls out of 4 = C(4, 2) = 4! \/ (2! * 2!) = (4 \u00d7 3) \/ (2 \u00d7 1) = 6. The number of favourable outcomes (both balls of the same colour) = Number of ways (both green) + Number of ways (both red) = 10 + 6 = 16. The probability that both balls are of the same colour is (Favourable outcomes) \/ (Total outcomes) = 16 \/ 36. Simplifying the fraction, 16\/36 = (4 \u00d7 4) \/ (9 \u00d7 4) = 4\/9.","robots":{"index":"index","follow":"follow","max-snippet":"max-snippet:-1","max-image-preview":"max-image-preview:large","max-video-preview":"max-video-preview:-1"},"canonical":"https:\/\/exam.pscnotes.com\/mcq\/from-a-pack-of-5-green-balls-and-4-red-balls-2-balls-are-drawn-at-ran\/","og_locale":"en_US","og_type":"article","og_title":"From a pack of 5 green balls and 4 red balls, 2 balls are drawn at ran","og_description":"The correct answer is 4\/9. Total number of balls in the pack is 5 green + 4 red = 9 balls. We are drawing 2 balls at random. The total number of ways to choose 2 balls out of 9 is given by the combination formula C(n, k) = n! \/ (k! * (n-k)!): Total outcomes = C(9, 2) = 9! \/ (2! * 7!) = (9 \u00d7 8) \/ (2 \u00d7 1) = 36. We want the probability that both balls are of the same colour. This can happen in two mutually exclusive ways: Case 1: Both balls are green. Number of ways to choose 2 green balls out of 5 = C(5, 2) = 5! \/ (2! * 3!) = (5 \u00d7 4) \/ (2 \u00d7 1) = 10. Case 2: Both balls are red. Number of ways to choose 2 red balls out of 4 = C(4, 2) = 4! \/ (2! * 2!) = (4 \u00d7 3) \/ (2 \u00d7 1) = 6. The number of favourable outcomes (both balls of the same colour) = Number of ways (both green) + Number of ways (both red) = 10 + 6 = 16. The probability that both balls are of the same colour is (Favourable outcomes) \/ (Total outcomes) = 16 \/ 36. Simplifying the fraction, 16\/36 = (4 \u00d7 4) \/ (9 \u00d7 4) = 4\/9.","og_url":"https:\/\/exam.pscnotes.com\/mcq\/from-a-pack-of-5-green-balls-and-4-red-balls-2-balls-are-drawn-at-ran\/","og_site_name":"MCQ and Quiz for Exams","article_published_time":"2025-06-01T11:42:30+00:00","author":"rawan239","twitter_card":"summary_large_image","twitter_misc":{"Written by":"rawan239","Est. reading time":"1 minute"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"WebPage","@id":"https:\/\/exam.pscnotes.com\/mcq\/from-a-pack-of-5-green-balls-and-4-red-balls-2-balls-are-drawn-at-ran\/","url":"https:\/\/exam.pscnotes.com\/mcq\/from-a-pack-of-5-green-balls-and-4-red-balls-2-balls-are-drawn-at-ran\/","name":"From a pack of 5 green balls and 4 red balls, 2 balls are drawn at ran","isPartOf":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#website"},"datePublished":"2025-06-01T11:42:30+00:00","dateModified":"2025-06-01T11:42:30+00:00","author":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209"},"description":"The correct answer is 4\/9. Total number of balls in the pack is 5 green + 4 red = 9 balls. We are drawing 2 balls at random. The total number of ways to choose 2 balls out of 9 is given by the combination formula C(n, k) = n! \/ (k! * (n-k)!): Total outcomes = C(9, 2) = 9! \/ (2! * 7!) = (9 \u00d7 8) \/ (2 \u00d7 1) = 36. We want the probability that both balls are of the same colour. This can happen in two mutually exclusive ways: Case 1: Both balls are green. Number of ways to choose 2 green balls out of 5 = C(5, 2) = 5! \/ (2! * 3!) = (5 \u00d7 4) \/ (2 \u00d7 1) = 10. Case 2: Both balls are red. Number of ways to choose 2 red balls out of 4 = C(4, 2) = 4! \/ (2! * 2!) = (4 \u00d7 3) \/ (2 \u00d7 1) = 6. The number of favourable outcomes (both balls of the same colour) = Number of ways (both green) + Number of ways (both red) = 10 + 6 = 16. The probability that both balls are of the same colour is (Favourable outcomes) \/ (Total outcomes) = 16 \/ 36. Simplifying the fraction, 16\/36 = (4 \u00d7 4) \/ (9 \u00d7 4) = 4\/9.","breadcrumb":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/from-a-pack-of-5-green-balls-and-4-red-balls-2-balls-are-drawn-at-ran\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/exam.pscnotes.com\/mcq\/from-a-pack-of-5-green-balls-and-4-red-balls-2-balls-are-drawn-at-ran\/"]}]},{"@type":"BreadcrumbList","@id":"https:\/\/exam.pscnotes.com\/mcq\/from-a-pack-of-5-green-balls-and-4-red-balls-2-balls-are-drawn-at-ran\/#breadcrumb","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https:\/\/exam.pscnotes.com\/mcq\/"},{"@type":"ListItem","position":2,"name":"UPSC CISF-AC-EXE","item":"https:\/\/exam.pscnotes.com\/mcq\/category\/upsc-cisf-ac-exe\/"},{"@type":"ListItem","position":3,"name":"From a pack of 5 green balls and 4 red balls, 2 balls are drawn at ran"}]},{"@type":"WebSite","@id":"https:\/\/exam.pscnotes.com\/mcq\/#website","url":"https:\/\/exam.pscnotes.com\/mcq\/","name":"MCQ and Quiz for Exams","description":"","potentialAction":[{"@type":"SearchAction","target":{"@type":"EntryPoint","urlTemplate":"https:\/\/exam.pscnotes.com\/mcq\/?s={search_term_string}"},"query-input":"required name=search_term_string"}],"inLanguage":"en-US"},{"@type":"Person","@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209","name":"rawan239","image":{"@type":"ImageObject","inLanguage":"en-US","@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/image\/","url":"https:\/\/secure.gravatar.com\/avatar\/761a7274f9cce048fa5b921221e7934820d74514df93ef195a9d22af0c1c9001?s=96&d=mm&r=g","contentUrl":"https:\/\/secure.gravatar.com\/avatar\/761a7274f9cce048fa5b921221e7934820d74514df93ef195a9d22af0c1c9001?s=96&d=mm&r=g","caption":"rawan239"},"sameAs":["https:\/\/exam.pscnotes.com"],"url":"https:\/\/exam.pscnotes.com\/mcq\/author\/rawan239\/"}]}},"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts\/93132","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/comments?post=93132"}],"version-history":[{"count":0,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts\/93132\/revisions"}],"wp:attachment":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/media?parent=93132"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/categories?post=93132"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/tags?post=93132"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}