Answer to Question #300857 in Discrete Mathematics for Tege

Question

4. An unbiased coin is tossed twice. If A is the event: both head or tail have occurred and B is the event: at most one tail is observed find : i) P(A) ii) P(B) iii) P(A\B) iv) P(B\A)

Answer 1

We define the sample space S = {HH, HT, TH, TT}

i) If A is the event that both head or tail have occurred, then the outcome of A is { HT, TH }

Thus P(A) = n(A) / n (S) = 2/4 = 1/2

II) Given B is the event: at most one tail is observed, then the outcomes will be { HH, TH, HT}

Thus P(B) = n(B) / n(S) = 3/4

III) We define  P(A\B) = P(A ∩ B ) / P(B)

Now A ∩ B has the outcomes { HT, TH }, Thus P(A ∩ B ) = 2/4 = 1/2

Hence  P(A\B) = (1/2) / (3/4) = 2/3

iv)  P(B\A) = P(A ∩ B ) / P(A)

Since P(A ∩ B ) = 1/2 and P(A) = 1/2

Then  P(B\A) = (1/2) / (1/2) = 1