anonymous
  • anonymous
I need help with simple random variables & binomial distributions, probabilities please :)
Mathematics
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anonymous
  • anonymous
I need help with simple random variables & binomial distributions, probabilities please :)
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
A 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
  • anonymous
Ok, 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
  • anonymous
Im supposed to be using binomial probability formula

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anonymous
  • anonymous
P(at least one correct)=1-P(all wrong)=\[1-10C4\times (3/4)^4\]
anonymous
  • anonymous
sorry I think it would be \[1-\left( \frac{3}{4} \right)^4\]
anonymous
  • anonymous
where do you get the upper four? i understand that 3/4 is p(wrong)
anonymous
  • anonymous
because i was thinking that its q up n-x
anonymous
  • anonymous
10- 9 right? to be at least 1?
anonymous
  • anonymous
answer is 1-(3/4)^10
anonymous
  • anonymous
\[\ the\ required\ probability\ is=10C1\times(1/4)^1\times(3/4)^9+10C2\times(1/4)^2\times(3/4)^8+...\]
anonymous
  • anonymous
= .94 your good, really good!
anonymous
  • anonymous
I am very sorry for the previuos wrong hasty answers by me. By the way if you have any problems regarding this ask me.
anonymous
  • anonymous
i 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
  • anonymous
I guess, what i am really asking is why did you add them
anonymous
  • anonymous
why 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
  • anonymous
ohhhhhhhhh i have to do this for EACH event? so I should probably start by listing all events possible
anonymous
  • anonymous
I 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
  • anonymous
ask if you have residual doubts.
anonymous
  • anonymous
like P{X=1) P(X=2) P(X=3) ? those are considered events right?
anonymous
  • anonymous
in 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
  • anonymous
But 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
  • anonymous
wow, 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
  • anonymous
thnx for the medal! :)
anonymous
  • anonymous
you deserved it!

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