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anonymous
 5 years ago
I need help with simple random variables & binomial distributions, probabilities please :)
anonymous
 5 years ago
I need help with simple random variables & binomial distributions, probabilities please :)

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0A random variable or stochastic variable is, roughly speaking, a variable whose value results from a measurement on some type of random process. The binomial distribution is the discrete probability distribution of the number of successes in a sequence of n independent yes/no experiments, each of which yields success with probability p. Such a success/failure experiment is also called a Bernoulli experiment or Bernoulli trial. Probability is a way of expressing knowledge or belief that an event will occur or has occurred.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ok, I am having a hard time setting the one up. A student takes a multiple choice exam with 10 questions, each with four possible selections for the answer. A passing grade is 60% or better. Suppose the student guesses at each question. Find the probability the student gets at least one question correct. I know what the answer is, I just can't figure out how to get to it.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Im supposed to be using binomial probability formula

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0P(at least one correct)=1P(all wrong)=\[110C4\times (3/4)^4\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sorry I think it would be \[1\left( \frac{3}{4} \right)^4\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0where do you get the upper four? i understand that 3/4 is p(wrong)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0because i was thinking that its q up nx

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.010 9 right? to be at least 1?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\ the\ required\ probability\ is=10C1\times(1/4)^1\times(3/4)^9+10C2\times(1/4)^2\times(3/4)^8+...\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0= .94 your good, really good!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I am very sorry for the previuos wrong hasty answers by me. By the way if you have any problems regarding this ask me.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i missed like a whole nother step. how come we do it twice? and so forth. i dont get how u plugged it in like that

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I guess, what i am really asking is why did you add them

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0why I added them? Its beacause of the sum rule of probabilty. putting more intuitively, each addend gives a way of completing the even. so we sum all the events

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ohhhhhhhhh i have to do this for EACH event? so I should probably start by listing all events possible

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I m sorry i said "we sum all the events". It should rather be "we sum all the ways of completing/doing the event". Yes, we should sum all the ways of a certain event. Here we are working with a single event. You should start listing all the possible ways of completing the events. am i clear?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ask if you have residual doubts.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0like P{X=1) P(X=2) P(X=3) ? those are considered events right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0in binomial distributions, we have to find the probabilty of a SET OF EVENTS. So P(x=1,2,3 etc) are EVENTS. You are completely right. But you can view the "small" events like p(x=1) etc. as ways of achieving the MEGA EVENT! in our question the mega event is "the student gets at least one right". Whenever there is a usage of words like "atleast/atmost" be sure that you have to find the prob. of a set of events OR a MEGAEVENT! (i named it just for FUN!!:) )so u consider all the events that can constitute the megaevent.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0But don't forget that each P(x=i where i=1,2,3...) is an event and can be accomplished in a number of ways, that is more than one way

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wow, seriously thank you so much your time. You saved me hours! And there are still 4 more parts to this story problem, so I'm going to try them out. Have a great night :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thnx for the medal! :)
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