An investor buys 100 shares of AT&T stock and records its price change daily. [a] List several possible events for this experiment. [b] Estimate the probability for each event you described in [a]. [c] [EXTRA CREDIT ] Which concept of probability did you use in [b]? [18] The events X and Y are mutually exclusive. Suppose P(X) = 0.05 and P(Y) = 0.02. [a] What is the probability of either X or Y occurring? [b] [EXTRA CREDIT] What is the probability that neither X nor Y will happen? [19] All Seasons Plumbing has two service trucks which frequently break down. If the probability the first truck is available is 0.75; the probability the second truck is available is 0.50; the probability that both are available is 0.30. What is the probability that neither truck is available? [20] [EXTRA CREDIT] If you ask three strangers about their birthdays, what is the probability that: [a] All were born on a Wednesday? [b] All were born on different days of the week? [c] None were born<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n [a] Possible Events for This Experiment<\/strong><\/p>\n\n\n\n [b] Estimating Probabilities<\/strong> [c] Concept of Probability Used<\/strong> [a] Probability of Either X or Y Occurring<\/strong> [b] Probability That Neither X nor Y Will Happen<\/strong> The probability that neither truck is available<\/strong> is: [a] All Born on a Wednesday<\/strong> [b] All Born on Different Days of the Week<\/strong> [c] None Born on a Wednesday<\/strong> Probability helps us predict the likelihood of outcomes in uncertain situations. For the AT&T stock, we used empirical probability<\/strong>, which relies on historical data, to estimate the chances of price changes. Mutually exclusive events, like X and Y, cannot happen simultaneously, so their probabilities add directly to find ( P(X \\cup Y) ). The complement rule ensures the sum of all possible outcomes equals 1.<\/p>\n\n\n\n For the trucks’ problem, joint probabilities account for overlapping scenarios, like both trucks being available, and subtraction ensures accuracy when combining probabilities.<\/p>\n\n\n\n The birthday problem is solved using the classical probability<\/strong> approach, assuming equal likelihood for any day of the week. Independent probabilities multiply for events like all being born on Wednesday. Constraints, such as avoiding overlaps (all different days), reduce options sequentially. Complementary probabilities (none on Wednesday) focus on what doesn\u2019t occur, reinforcing the completeness of probabilities.<\/p>\n\n\n\n This logical framework enables accurate predictions, demonstrating probability’s vital role in decision-making across disciplines.<\/p>\n\n\n\n <\/p>\n","protected":false},"excerpt":{"rendered":" An investor buys 100 shares of AT&T stock and records its price change daily. [a] List several possible events for this experiment. [b] Estimate the probability for each event you described in [a]. [c] [EXTRA CREDIT ] Which concept of probability did you use in [b]? [18] The events X and Y are mutually exclusive. […]<\/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-183071","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/183071","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=183071"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/183071\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=183071"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=183071"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=183071"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Solutions and Explanations<\/strong><\/h3>\n\n\n\n
[17] AT&T Stock Price Experiment<\/strong><\/h4>\n\n\n\n
\n
Assuming historical stock data suggests the following probabilities:<\/p>\n\n\n\n\n
The empirical probability<\/strong> concept is used. Probabilities are estimated based on historical or past observations of AT&T stock price changes.<\/p>\n\n\n\n
\n\n\n\n[18] Mutually Exclusive Events X and Y<\/strong><\/h4>\n\n\n\n
For mutually exclusive events:
[
P(X \\cup Y) = P(X) + P(Y)
]
Substitute ( P(X) = 0.05 ) and ( P(Y) = 0.02 ):
[
P(X \\cup Y) = 0.05 + 0.02 = 0.07
]<\/p>\n\n\n\n
The complement rule applies:
[
P(\\text{Neither X nor Y}) = 1 – P(X \\cup Y)
]
Substitute ( P(X \\cup Y) = 0.07 ):
[
P(\\text{Neither X nor Y}) = 1 – 0.07 = 0.93
]<\/p>\n\n\n\n
\n\n\n\n[19] Availability of Trucks<\/strong><\/h4>\n\n\n\n
[
P(\\text{Neither}) = 1 – P(\\text{At least one is available})
]
Using the inclusion-exclusion principle for at least one being available:
[
P(\\text{At least one is available}) = P(\\text{First}) + P(\\text{Second}) – P(\\text{Both})
]
Substitute ( P(\\text{First}) = 0.75 ), ( P(\\text{Second}) = 0.50 ), and ( P(\\text{Both}) = 0.30 ):
[
P(\\text{At least one is available}) = 0.75 + 0.50 – 0.30 = 0.95
]
Now, compute ( P(\\text{Neither}) ):
[
P(\\text{Neither}) = 1 – 0.95 = 0.05
]<\/p>\n\n\n\n
\n\n\n\n[20] Birthdays<\/strong><\/h4>\n\n\n\n
The probability that one person is born on a specific day of the week (e.g., Wednesday) is ( \\frac{1}{7} ). For three independent people:
[
P(\\text{All on Wednesday}) = \\left( \\frac{1}{7} \\right)^3 = \\frac{1}{343} \\approx 0.0029
]<\/p>\n\n\n\n
The first person has 7 options, the second has 6 (to avoid the first’s day), and the third has 5:
[
P(\\text{All different days}) = \\frac{7}{7} \\cdot \\frac{6}{7} \\cdot \\frac{5}{7} = \\frac{210}{343} \\approx 0.612
]<\/p>\n\n\n\n
The probability one person is not born on Wednesday is ( \\frac{6}{7} ). For three independent people:
[
P(\\text{None on Wednesday}) = \\left( \\frac{6}{7} \\right)^3 = \\frac{216}{343} \\approx 0.63
]<\/p>\n\n\n\n
\n\n\n\n300-Word Explanation<\/strong><\/h3>\n\n\n\n