Answer:
[tex]P(HH)=\frac{13}{52} \frac{12}{51}=0.0588[/tex]
[tex]P(HN)=\frac{13}{52} \frac{39}{51}=0,1912[/tex]
[tex]P(NH)=\frac{39}{52} \frac{13}{51}=0,1912[/tex]
[tex]P(NN)=\frac{39}{52} \frac{38}{51}=0,5588[/tex]
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 [tex]\frac{13}{52}[/tex]. The second heart is taken from 9 available out of 51 cards. The probability is [tex]\frac{12}{51}[/tex]. The combined probability is the product of both
[tex]P(HH)=\frac{13}{52} \frac{12}{51}=0.0588[/tex]
Case 2: HN
The probability of H in the first deal is already computed: [tex]\frac{13}{52}[/tex]. 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 [tex]\frac{39}{51}[/tex]
[tex]P(HN)=\frac{13}{52} \frac{39}{51}=0,1912[/tex]
Case 3: NH
The probability of taking a non-heart as the first card is [tex]\frac{39}{52}[/tex]. The second card is a heart, taken from the remaining 13 with probability [tex]\frac{13}{51}[/tex]
[tex]P(NH)=\frac{39}{52} \frac{13}{51}=0,1912[/tex]
Case 4: NN
Similar reasoning leads to
[tex]P(NN)=\frac{39}{52} \frac{38}{51}=0,5588[/tex]
Notice all of them sum 1
