rrH SAMPLE For the following questions circle may not receive credit your answer

Question

rrH SAMPLE For the following questions circle may not receive credit your answer and show all work Solutions without work or reasoning 14. A multiple choice quiz has three questions, each with five answer choices. One correct. You have no idea what the answer is to any question and have to guess answer is each answer a) Find the probability of answering the first question correctly and all of the remaining questions incorrectly. b) Find the probability of answering none of the questions correctly. c) Find the probability of answering at least one question correctly. 15. (5 points) If a single fair 6 sided die is rolled 3 times, what is the probability it will land on a 3, then on a 2, and then on a 3 again? 3. suces dependent

Explanation / Answer

14.
BIONOMIAL DISTRIBUTION
pmf of B.D is = f ( k ) = ( n k ) p^k * ( 1- p) ^ n-k
where
k = number of successes in trials   
n = is the number of independent trials   
p = probability of success on each trial
I.
mean = np
where
n = total number of repetitions experiment is excueted
p = success probability
mean = 3 * 0.2
= 0.6
II.
variance = npq
where
n = total number of repetitions experiment is excueted
p = success probability
q = failure probability
variance = 3 * 0.2 * 0.8
= 0.48
III.
standard deviation = sqrt( variance ) = sqrt(0.48)
=0.6928
a.
P( X = 1 ) = ( 3 1 ) * ( 0.2^1) * ( 1 - 0.2 )^2
= 0.384
b.
P( X = 0 ) = ( 3 0 ) * ( 0.2^0) * ( 1 - 0.2 )^3
= 0.512
c.
P( X < 1) = P(X=0)
= ( 3 0 ) * 0.2^0 * ( 1- 0.2 ) ^3
= 0.512
P( X > = 1 ) = 1 - P( X < 1) = 0.488
15.
BIONOMIAL DISTRIBUTION
pmf of B.D is = f ( k ) = ( n k ) p^k * ( 1- p) ^ n-k
where
k = number of successes in trials   
n = is the number of independent trials   
p = probability of success on each trial
I.
mean = np
where
n = total number of repetitions experiment is excueted
p = success probability
mean = 3 * 0.166
= 0.498
II.
variance = npq
where
n = total number of repetitions experiment is excueted
p = success probability
q = failure probability
variance = 3 * 0.166 * 0.834
= 0.4153
III.
standard deviation = sqrt( variance ) = sqrt(0.4153)
=0.6445
a.
P( X = 3 ) = ( 3 3 ) * ( 0.166^3) * ( 1 - 0.166 )^0
= 0.0046
b.
P( X = 2 ) = ( 3 2 ) * ( 0.166^2) * ( 1 - 0.166 )^1
= 0.0689
c.
P( X = 3 ) = ( 3 3 ) * ( 0.166^3) * ( 1 - 0.166 )^0
= 0.0046

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