A student guesses the answers to 6 questions on a true-false quiz. Find the probability that the indicated number of guesses are correct: no more than 2(hint:no more than 2 means exactly 0 means exactly 1 or exactly 2)
so you have to compute \[P(X=0), P(X=1), P(X=2)\] where \(X\) is the number of correct responses binomial for this one
the probability of guessing correctly is the same as the probability of guessing incorrectly both \(\frac{1}{2}\) so \[P(X=0)=\left(\frac{1}{2}\right)^6\]
X = # of right answers of 6 guesses (true/false) X is a random binomial variable with n=6, p = .5 we want no more than 2 , so P( X <= 2 ) calculator says binomcdf ( 6, .5, 2 ) =
Let X = # of right answers of 6 guesses (true/false) X is a random binomial variable with n=6, p = .5 we want no more than 2 , so P( X <= 2 ) calculator says binomcdf ( 6, .5, 2 ) = .34375 = 11/32
\[P(X=1)=\binom{6}{1}\left(\frac{1}{2}\right)^6=6\left(\frac{1}{2}\right)^6\]
Let X = # of right answers of 6 guesses (true/false) X is a random binomial variable with n=6, p = .5 we want no more than 2 which is 2 or less P( X <= 2 ) = P( X=0 or X = 1 or X=2) = P(X=0) + P(X=1) + P(X=2) calculator says binomcdf ( 6, .5, 2 ) = .34375 = 11/32
there are 8 cars i a parking lot on very cold day suppose the probability of any of them not starting is 0.31 what is the probability that exactly 2 of the cars will not start
Let X = # of cars not starting X is a random binomial variable with n=6, p = .31 we want P ( X = 2) calculator says binompdf ( 8, .31, 2 )
.209
205?
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