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.
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
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
To determine the probabilities, we first need to understand the total number of possible outcomes and the number of favorable outcomes for each event.
1. Probability that at least one card is an ace:
Let's first determine the probability that neither card is an ace. There are 52 cards in a deck, and 4 of them are aces. So, the probability that one card is not an ace is 48/52, and for both cards, it would be (48/52) * (48/52).
Therefore, the probability that both cards are not aces is (48/52) * (48/52). To find the probability that at least one card is an ace, we need to subtract this probability from 1. So, the final probability is 1 - [(48/52) * (48/52)].
Simplifying it further, the probability that at least one card is an ace is 1 - (48/52)^2.
2. Probability that the two cards are of the same suit:
There are 4 suits in a deck, and for each suit, we have 13 cards. So, the probability of selecting a card of a specific suit is 13/52. Since there are four suits, the probability of drawing two cards of the same suit is (13/52) * (13/52).
3. Probability that neither card is an ace:
As mentioned earlier, there are 4 aces in a deck of 52 cards. So, the probability that a card is not an ace is 1 - (4/52), which simplifies to 48/52. Similar to the above case, we take the probability for both cards, which is (48/52) * (48/52).
4. Probability that neither card is a diamond or club:
In a deck of 52 cards, there are 13 diamonds and 13 clubs. So, the probability that a card is neither a diamond nor a club is 1 - [(13/52) + (13/52)], which simplifies to 26/52.
To summarize:
1. The probability that at least one card is an ace is 1 - (48/52)^2.
2. The probability that the two cards are of the same suit is (13/52) * (13/52).
3. The probability that neither card is an ace is (48/52) * (48/52).
4. The probability that neither card is a diamond or a club is 26/52.