Two cards are dealt from the top of a well-shuffled standard deck of cards.

a. draw a tree diagram showing the probabilities of a heart (h) and non-heart (n) for each of the two cards.

b. find p(hh), p(hn), p(nh), and p(nn).

(only need with b! )

Step-by-step explanation:

Probabilities with events without replacement

When a random experience is performed without replacement, it means the conditions for the first and subsequent observations are not the same, the sample space changes. The throw of a die is an example of a random experience with replacement since each throw is done with the same die in the same conditions. Taking two cards from a deck of cards is a non-replacing experience.

b)

In the random experience described in the question, two cards are dealt from a deck of cards. The first card is taken from a deck of 52 cards, the second from a deck of 51 cards. We are observing if the card is a Heart (H) or a non-heart (N). Four things can happen:  

HH -> Two hearts are dealt

HN -> First card is a Heart, the second card is a non-heart

NH -> First card is a non-heart, the second card is a Heart

NN -> Two non-hearts are dealt

Case 1: HH

The first Heart is taken from 10 available. The probability of that event is . The second heart is taken from 9 available out of 51 cards. The probability is . The combined probability is the product of both

Case 2: HN

The probability of H in the first deal is already computed: . The second card is a non-heart, which can be taken from all the non-heart cards available, 39 in total. So the probability of taking a non-heart is  

Case 3: NH

The probability of taking a non-heart as the first card is . The second card is a heart, taken from the remaining 13 with probability

Case 4: NN

Similar reasoning leads to

Notice all of them sum 1

it would be 4 over -2 divided by -1 over 3.

if you want a solution it would be 4

you flip it so it would be 4/-2 divided by 3/-1 then it would be 12/3 and you would simplify it to 4