Quiz 13
Discrete Mathematics · 2 problems · solutions hidden, click to reveal
Flashcards
Problems
A fair die is rolled twice. Let \(A\) be "the sum is 5" and \(B\) be "the first roll is even."
(a) Find \(p(A)\). (b) Find \(p(A \mid B)\). (c) Are \(A\) and \(B\) independent?
(a) Find \(p(A)\). (b) Find \(p(A \mid B)\). (c) Are \(A\) and \(B\) independent?
Show solution
\(A = \{(1,4),(2,3),(3,2),(4,1)\}\), so \(p(A) = 4/36 = 1/9\).
\(A \cap B = \{(2,3),(4,1)\}\), \(p(A\cap B) = 2/36 = 1/18\), and \(p(B) = 1/2\), so \(p(A\mid B) = \dfrac{1/18}{1/2} = 1/9\).
Since \(p(A\mid B) = p(A)\), \(A\) and \(B\) are independent.
\(A \cap B = \{(2,3),(4,1)\}\), \(p(A\cap B) = 2/36 = 1/18\), and \(p(B) = 1/2\), so \(p(A\mid B) = \dfrac{1/18}{1/2} = 1/9\).
Since \(p(A\mid B) = p(A)\), \(A\) and \(B\) are independent.
A biased coin has heads twice as likely as tails. Flipped three times, what is the probability of at least one tail?
Show solution
Answer: \(19/27\).
\(p(H) = 2/3,\ p(T) = 1/3\). Using the complement, \(p(\text{at least one tail}) = 1 - p(HHH) = 1 - (2/3)^3 = 1 - 8/27 = 19/27.\)