Question

You roll a fair die 5 times. What is the probability you get at least four 6’s? This time you roll the die 204 times. What is the probability you get between 30 and 40 6’s?

Answer #1

Binomial distribution: P(X) = nCx p^{x}
q^{n-x}

P(at least four 6's in 5 rolls) = P(4) + P(5) + P(6)

= 6C4x(1/6)^{4}(5/6)^{2} +
6C5x(1/6)^{5}x(5/6) + (1/6)^{6}

= 0.0080 + 0.0006 + 0.0001

= 0.0087

Norml approximation for binomial distribution

n = 204

p = 1/6

q = 5/6

Mean = np = 34

Standard deviation = = 5.323

P(X < A) = P(Z < (A - mean)/standard deviation)

P(between 30 and 40 6’s) = P(X < 40.5) - P(X < 29.5) (continuity correction of 0.5 applied)

= P(Z < (40.5 - 34)/5.323) - P(Z < (29.5 - 34)/5.323)

= P(Z < 1.22) - P(Z < -0.85)

= 0.8888 - 0.1977

= 0.6911

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1. Game of rolling dice
a. Roll a fair die once. What is the sample space? What is the
probability to get “six”? What is the probability to get “six” or
“five”?
b. Roll a fair die 10 times. What is the probability to get
“six” twice? What is the probability to get six at
least twice?
c. Roll a fair die 10 times. What is the expected value and
variance of getting “six”?
d. If you roll the die...

You roll a fair 6 sided die 5 times. Let Xi be the
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(b) What is Cov(X1, X2)?
(c) Given that X1 = 2, what is the probability the
first roll is a 1?
(d) Given that X1 = 2, what is the conditional probability mass
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X2?
(e) What is E[X2|X1]

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Group of answer choices

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Please Explain!
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