prob no ace = 1 - 4/52 = 12/13
1. so we don't want (no ace, no ace)
prob of stated event = 1 - (12/13)^2 = 25/169
2. could be SS, CC, HH, or DD
prob(spade, spade) = (4/52)^2 = 1/169
same for each of the others, so
prob of stated event = 4/169
3. prob(no ace, no ace) = (12/13)^2 = 144/169
4. you try the last one, and let me know what you got with explanation
Anne and Bob each have a deck of playing cards. Each flips over a randomly selected card. Assume
that all pairs of cards are equally likely to be drawn. Determine the following probabilities: (1)
The probability that at least one card is an ace, (2) the probability that the two cards are of the
same suit, (3) the probability that neither card is an ace, and (4) the probability that neither card
is a diamond or club.
2 answers
A coin is such that the tail is four times as likely as the head. A game is played such that you earn 4 points for a head and lose 2 points for a tail after every toss. Let X be the total score after 3 consecutive tosses. Find
(i) the probability distribution of X
(ii) and interpret expected value of X
(iii) and interpret variance of X
(i) the probability distribution of X
(ii) and interpret expected value of X
(iii) and interpret variance of X