{"id":207956,"date":"2025-04-25T14:42:45","date_gmt":"2025-04-25T14:42:45","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=207956"},"modified":"2025-04-25T14:42:47","modified_gmt":"2025-04-25T14:42:47","slug":"consider-randomly-selecting-a-student-at-a-large-university","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/04\/25\/consider-randomly-selecting-a-student-at-a-large-university\/","title":{"rendered":"Consider randomly selecting a student at a large university"},"content":{"rendered":"\n<p>Consider randomly selecting a student at a large university, and let A be the event that the selected student has a Visa card and B be the analogous event for MasterCard. Suppose that P(A) = .6 and P(B) = .<\/p>\n\n\n\n<p>a. Could it be the case that P(A &gt; B) = .5? Why or why not?<\/p>\n\n\n\n<p>b. From now on, suppose that P(A &gt; B) = .3. What is the probability that the selected student has at least one of these two types of cards?<\/p>\n\n\n\n<p>c. What is the probability that the selected student has neither type of card?<\/p>\n\n\n\n<p>d. Describe, in terms of A and B, the event that the selected student has a Visa card but not a MasterCard, and then calculate the probability of this event.<\/p>\n\n\n\n<p>e. Calculate the probability that the selected student has exactly one of the two types of cards.<\/p>\n\n\n\n<p><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-6-color\"><strong>The correct answer and explanation is :<\/strong><\/mark><\/p>\n\n\n\n<p>To address the given questions, we need to use the probability theory and basic set operations. Let\u2019s break down each part and work through the solution step-by-step.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Given Information:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>P(A)<\/strong> = Probability that a student has a Visa card = 0.6<\/li>\n\n\n\n<li><strong>P(B)<\/strong> = Probability that a student has a MasterCard (this is unspecified, so we\u2019ll assume this is provided elsewhere in the full problem or is intended to be determined).<\/li>\n\n\n\n<li><strong>P(A \u2229 B)<\/strong> = Probability that a student has both cards. This is not explicitly given, so we&#8217;ll need to work with this assumption and the other information provided.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">a. Could it be the case that P(A &gt; B) = 0.5? Why or why not?<\/h3>\n\n\n\n<p><strong>P(A &gt; B)<\/strong> typically refers to the probability that event A (having a Visa card) occurs, given that event B (having a MasterCard) does not occur, or some comparison between the two events. However, based on the context of probabilities, <strong>P(A &gt; B)<\/strong> doesn&#8217;t represent a valid probability expression as it is currently written. Probabilities need to be between 0 and 1, and what we usually compare in probability theory is <strong>P(A \u2229 B)<\/strong> (both events occurring), <strong>P(A \u222a B)<\/strong> (either event occurring), or conditional probabilities (e.g., P(A | B)).<\/p>\n\n\n\n<p>Without clear clarification of <strong>P(A &gt; B)<\/strong>, we can reasonably conclude that this statement as written doesn\u2019t reflect a standard probability expression. Therefore, <strong>P(A &gt; B) = 0.5<\/strong> is likely not a correct or possible interpretation.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">b. If <strong>P(A &gt; B) = 0.3<\/strong>, what is the probability that the selected student has at least one of these two types of cards?<\/h3>\n\n\n\n<p>We need to calculate <strong>P(A \u222a B)<\/strong>, the probability that the student has at least one of the two cards. This can be found using the formula for the union of two events:<br>[<br>P(A \\cup B) = P(A) + P(B) &#8211; P(A \\cap B)<br>]<\/p>\n\n\n\n<p>We know <strong>P(A) = 0.6<\/strong>, and we are told that <strong>P(A &gt; B) = 0.3<\/strong>, which typically refers to the conditional probability that a student has a Visa card but not a MasterCard. Therefore, <strong>P(A \u2229 B&#8217;) = 0.3<\/strong>, where <strong>B&#8217;<\/strong> is the complement of <strong>B<\/strong> (i.e., the student does not have a MasterCard).<\/p>\n\n\n\n<p>Since <strong>P(A \u2229 B&#8217;)<\/strong> is the probability that the student has a Visa but not a MasterCard, we can use:<br>[<br>P(A) = P(A \\cap B) + P(A \\cap B&#8217;)<br>]<br>[<br>0.6 = P(A \\cap B) + 0.3<br>]<br>Thus, <strong>P(A \u2229 B) = 0.3<\/strong>.<\/p>\n\n\n\n<p>Now we can substitute into the formula for the union:<br>[<br>P(A \\cup B) = 0.6 + P(B) &#8211; 0.3<br>]<br>To fully compute <strong>P(A \u222a B)<\/strong>, we need the value of <strong>P(B)<\/strong>, which isn&#8217;t provided directly in the problem. However, if <strong>P(B) = 0.6<\/strong> (assuming equal probability for Visa and MasterCard holders):<br>[<br>P(A \\cup B) = 0.6 + 0.6 &#8211; 0.3 = 0.9<br>]<br>Therefore, the probability that the student has at least one of these two types of cards is <strong>0.9<\/strong>.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">c. What is the probability that the selected student has neither type of card?<\/h3>\n\n\n\n<p>The probability that the student has neither card is simply the complement of the probability that the student has at least one of the cards:<br>[<br>P(\\text{neither}) = 1 &#8211; P(A \\cup B)<br>]<br>Using the previously computed value for <strong>P(A \u222a B)<\/strong>:<br>[<br>P(\\text{neither}) = 1 &#8211; 0.9 = 0.1<br>]<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">d. Describe, in terms of A and B, the event that the selected student has a Visa card but not a MasterCard, and then calculate the probability of this event.<\/h3>\n\n\n\n<p>The event that the student has a Visa card but not a MasterCard is represented by <strong>A \u2229 B&#8217;<\/strong> (Visa but not MasterCard). We are given that:<br>[<br>P(A \\cap B&#8217;) = 0.3<br>]<br>So, the probability that the student has a Visa card but not a MasterCard is <strong>0.3<\/strong>.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">e. Calculate the probability that the selected student has exactly one of the two types of cards.<\/h3>\n\n\n\n<p>The probability that the student has exactly one of the two cards is the sum of the probabilities of having only a Visa or only a MasterCard. This is:<br>[<br>P(\\text{exactly one card}) = P(A \\cap B&#8217;) + P(B \\cap A&#8217;)<br>]<br>We already know <strong>P(A \u2229 B&#8217;) = 0.3<\/strong>, and similarly, <strong>P(B \u2229 A&#8217;)<\/strong> (having only a MasterCard) can be calculated as:<br>[<br>P(B \\cap A&#8217;) = P(B) &#8211; P(A \\cap B) = 0.6 &#8211; 0.3 = 0.3<br>]<br>Therefore, the probability of having exactly one card is:<br>[<br>P(\\text{exactly one card}) = 0.3 + 0.3 = 0.6<br>]<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Final Answers:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>a. <strong>P(A > B) = 0.5<\/strong> is not a valid probability expression.<\/li>\n\n\n\n<li>b. <strong>P(A \u222a B) = 0.9<\/strong> (assuming P(B) = 0.6).<\/li>\n\n\n\n<li>c. <strong>P(neither) = 0.1<\/strong>.<\/li>\n\n\n\n<li>d. <strong>P(A \u2229 B&#8217;) = 0.3<\/strong>.<\/li>\n\n\n\n<li>e. <strong>P(exactly one card) = 0.6<\/strong>.<\/li>\n<\/ul>\n","protected":false},"excerpt":{"rendered":"<p>Consider randomly selecting a student at a large university, and let A be the event that the selected student has a Visa card and B be the analogous event for MasterCard. Suppose that P(A) = .6 and P(B) = . a. Could it be the case that P(A &gt; B) = .5? Why or why [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-207956","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/207956","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=207956"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/207956\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=207956"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=207956"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=207956"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}