We are given the probability distribution of a discrete random variable XXX as:<\/p>\n\n\n\n
- \n
- f(0)=P(X=0)=0.89f(0) = P(X = 0) = 0.89f(0)=P(X=0)=0.89<\/li>\n\n\n\n
- f(1)=P(X=1)=0.07f(1) = P(X = 1) = 0.07f(1)=P(X=1)=0.07<\/li>\n\n\n\n
- f(2)=P(X=2)=0.03f(2) = P(X = 2) = 0.03f(2)=P(X=2)=0.03<\/li>\n\n\n\n
- f(3)=P(X=3)=0.01f(3) = P(X = 3) = 0.01f(3)=P(X=3)=0.01<\/li>\n<\/ul>\n\n\n\n
We are asked to find:P(0<X<2)+P(X=3)P(0 < X < 2) + P(X = 3)P(0<X<2)+P(X=3)<\/p>\n\n\n\n
Step-by-step Solution:<\/h3>\n\n\n\n
Step 1: Understand the expression P(0<X<2)P(0 < X < 2)P(0<X<2)<\/h4>\n\n\n\n
Since XXX is a discrete random variable, P(0<X<2)P(0 < X < 2)P(0<X<2) means the probability that XXX is greater than 0 but less than 2. The only integer value that satisfies this condition is X=1X = 1X=1. Therefore,P(0<X<2)=P(X=1)=0.07P(0 < X < 2) = P(X = 1) = 0.07P(0<X<2)=P(X=1)=0.07<\/p>\n\n\n\n
Step 2: Add the probability P(X=3)P(X = 3)P(X=3)<\/h4>\n\n\n\n
From the given distribution:P(X=3)=0.01P(X = 3) = 0.01P(X=3)=0.01<\/p>\n\n\n\n
Step 3: Add the two probabilities<\/h4>\n\n\n\n
P(0<X<2)+P(X=3)=0.07+0.01=0.08P(0 < X < 2) + P(X = 3) = 0.07 + 0.01 = 0.08P(0<X<2)+P(X=3)=0.07+0.01=0.08<\/p>\n\n\n\n
\n\n\n\nFinal Answer:<\/h3>\n\n\n\n
0.08\\boxed{0.08}0.08\u200b<\/p>\n\n\n\n
\n\n\n\nTextbook-style Explanation<\/h3>\n\n\n\n
In probability theory, a discrete probability distribution<\/strong> lists the probabilities associated with each possible value that a discrete random variable can take. In this case, the random variable XXX can take on the values 0, 1, 2, and 3, with the respective probabilities:<\/p>\n\n\n\n
- \n
- P(X=0)=0.89P(X = 0) = 0.89P(X=0)=0.89<\/li>\n\n\n\n
- P(X=1)=0.07P(X = 1) = 0.07P(X=1)=0.07<\/li>\n\n\n\n
- P(X=2)=0.03P(X = 2) = 0.03P(X=2)=0.03<\/li>\n\n\n\n
- P(X=3)=0.01P(X = 3) = 0.01P(X=3)=0.01<\/li>\n<\/ul>\n\n\n\n
We are interested in computing the sum of two probabilities: P(0<X<2)P(0 < X < 2)P(0<X<2) and P(X=3)P(X = 3)P(X=3). The expression 0<X<20 < X < 20<X<2 refers to the values of XXX that are greater than 0 and less than 2. Since XXX is a discrete variable, we only consider integer values. The only integer between 0 and 2 is 1. Therefore:P(0<X<2)=P(X=1)P(0 < X < 2) = P(X = 1)P(0<X<2)=P(X=1)<\/p>\n\n\n\n
From the given distribution, we find:P(X=1)=0.07andP(X=3)=0.01P(X = 1) = 0.07 \\quad \\text{and} \\quad P(X = 3) = 0.01P(X=1)=0.07andP(X=3)=0.01<\/p>\n\n\n\n
Adding these probabilities gives the final result:P(0<X<2)+P(X=3)=0.07+0.01=0.08P(0 < X < 2) + P(X = 3) = 0.07 + 0.01 = 0.08P(0<X<2)+P(X=3)=0.07+0.01=0.08<\/p>\n\n\n\n
This solution highlights the importance of carefully interpreting inequality expressions in the context of discrete random variables. Only integer values that satisfy the given conditions contribute to the probability sum.<\/p>\n\n\n\n
![\"\"](\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner7-412.jpeg\")
<\/figure>\n","protected":false},"excerpt":{"rendered":"The probability distribution of a random variable X is given by f(0) = 0.89, f(1) = 0.07, f(2) = 0.03, f(3) = 0.01. Then, P(0 < X < 2) + P(X = 3) = The Correct Answer and Explanation is: We are given the probability distribution of a discrete random variable XXX as: We are […]<\/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-232949","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/232949","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=232949"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/232949\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=232949"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=232949"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=232949"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}