Mathematics, 06.12.2019 05:31 straightbarz5916

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! )

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Answer from: sl1010

$P(HH)=\frac{13}{52} \frac{12}{51}=0.0588$

$P(HN)=\frac{13}{52} \frac{39}{51}=0,1912$

$P(NH)=\frac{39}{52} \frac{13}{51}=0,1912$

$P(NN)=\frac{39}{52} \frac{38}{51}=0,5588$

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.

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 $\frac{13}{52}$ . The second heart is taken from 9 available out of 51 cards. The probability is $\frac{12}{51}$ . The combined probability is the product of both

$P(HH)=\frac{13}{52} \frac{12}{51}=0.0588$

Case 2: HN

The probability of H in the first deal is already computed: $\frac{13}{52}$ . 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 $\frac{39}{51}$

$P(HN)=\frac{13}{52} \frac{39}{51}=0,1912$

Case 3: NH

The probability of taking a non-heart as the first card is $\frac{39}{52}$ . The second card is a heart, taken from the remaining 13 with probability $\frac{13}{51}$

$P(NH)=\frac{39}{52} \frac{13}{51}=0,1912$

Case 4: NN

Similar reasoning leads to

$P(NN)=\frac{39}{52} \frac{38}{51}=0,5588$

Notice all of them sum 1

Answer from: Quest

step-by-step explanation:

Answer from: Quest

the functions have the same rate of change i think. i narrowed it down to this answer and "the rate of change of tour 1 is greater" but i got it wrong using that answer so im assuming its the answer above. (this is a question i also answered from the iready site if anyone needs a reference)

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