"2.4.1 Discrete distributions and the binomial Flashcards","topicId":"c4456d42-08aa-4521-9def-2236b83e8c59","topicSlug":"discrete-distributions-and-the-binomial","cards":[{"id":"d735d7af-dd28-4ea8-afc2-88d19b323ef9","cardType":"basic","front":"What does a random variable assign to each outcome of a random process?","back":"A numerical value.","points":1,"mcqOptions":null},{"id":"1410a4b0-faed-48cb-9294-cb09c56bd275","cardType":"basic","front":"Why is the number of attempts needed to score a goal a discrete random variable?","back":"It takes only separate, countable values.","points":2,"mcqOptions":null},{"id":"ad700937-40bd-49a4-b8e3-9c6a4a5d57b5","cardType":"basic","front":"Why is height generally continuous rather than discrete?","back":"It can take any value within an interval.","points":2,"mcqOptions":null},{"id":"11fef51d-097f-4cc6-9910-e1d55adacfa6","cardType":"cloze","front":"For every possible value $x$, a valid distribution requires $\\text{[...]}$.","back":"For every possible value $x$, a valid distribution requires $\\boxed{0\\le P(X=x)\\le 1}$.","points":null,"mcqOptions":null},{"id":"d51bdc5e-bec6-4dda-bcfc-69365e77506b","cardType":"cloze","front":"Across all possible values, a valid distribution requires $\\text{[...]}$.","back":"Across all possible values, a valid distribution requires $\\boxed{\\sum P(X=x)=1}$.","points":null,"mcqOptions":null},{"id":"8531fd0f-b00c-47c4-8a41-d2ee4cc357ba","cardType":"basic","front":"A distribution has probabilities $0.2$, $k$, $0.35$ and $0.1$. Find $k$.","back":"$k=0.35$","points":1,"mcqOptions":null},{"id":"06ed0b49-f7b5-48da-9b69-d005779a00ae","cardType":"basic","front":"Are the possible values of a discrete random variable automatically equally likely?","back":"No - use or derive their probabilities.","points":1,"mcqOptions":null},{"id":"15f1da8e-dcf5-4761-9fec-8b829305767b","cardType":"basic","front":"What does $X\\sim B(n,p)$ state about $X$?","back":"$$X$ counts successes in $n$ trials, each with success probability $p$.","points":1,"mcqOptions":null},{"id":"075c1162-9716-473d-bbdb-f55c685b2bd5","cardType":"basic","front":"What is the probability of failure when the success probability is $p$?","back":"$1-p$","points":1,"mcqOptions":null},{"id":"67431331-c6be-465f-9ad5-4993cbd2fc7c","cardType":"basic","front":"What four conditions are required for a binomial model?","back":"

Fixed number of trials
Exactly two outcomes per trial
Constant success probability
Independent trials

","points":3,"mcqOptions":null},{"id":"896605be-e40a-47e3-8d95-b54bd0bcfa3a","cardType":"basic","front":"In a binomial model of defective components, why may a defect be called a success?","back":"A success is simply the outcome being counted.","points":2,"mcqOptions":null},{"id":"2163370c-2172-4f1d-b006-4791dabd8f45","cardType":"basic","front":"Why is sampling without replacement generally unsuitable for a binomial model?","back":"

Trials are usually not independent
The success probability changes

","points":3,"mcqOptions":null},{"id":"4737c87a-14e3-4550-b77c-cdc065deeaae","cardType":"cloze","front":"For $X\\sim B(n,p)$, $\\text{[...]}$.","back":"For $X\\sim B(n,p)$, $\\boxed{P(X=r)=\\binom{n}{r}p^r(1-p)^{n-r}}$.","points":null,"mcqOptions":null},{"id":"001a9cf3-d910-4621-bc8e-32a756743c04","cardType":"basic","front":"What does $\\binom{n}{r}$ count in the binomial probability formula?","back":"The ways to position $r$ successes among $n$ trials.","points":1,"mcqOptions":null},{"id":"231f90e1-170b-4ac5-adc1-3853a25b1b03","cardType":"cloze","front":"The combination coefficient is $\\text{[...]}$.","back":"The combination coefficient is $\\boxed{\\binom{n}{r}=\\frac{n!}{r!(n-r)!}}$.","points":null,"mcqOptions":null},{"id":"b3633ae6-0f68-40d5-aa92-9f2116d766ca","card