Respuesta :
(a) No. The probability of drawing a specific second card depends on the identity of the first card.
Two events are independent if knowledge on the first doesn't change anything about the probability distribution of the second event. This isn't the case: since you don't replace the first card, you have some extra knowledge about the second.
If, for example, you pick the 7 of hearts at the beginning, you are sure that the second card won't be the 7 of hearts. This extra piece of information depends on the first pick, and so the two pick are not independent: knowledge about the first pick changed what you expect from the second pick.
(b)
The probability of finding ace on 1st card is 4/52 = 1/13, because there are four aces (one for each suit) out of 52 cards. If this happens, you're left with 51 cards, 4 of which are kings (one for each suit). So, the probability of picking a king among the remaining cards is 4/51. The probability of the two happening one after the other is
[tex]\dfrac{1}{13}\cdot\dfrac{4}{51}=\dfrac{4}{663}[/tex]
(c)
You can use the same logic as point (b)
(d)
We have 8 "good" cards for the first pick: any of the 4 aces or any of the 4 kings. So, the first pick has a success rate of 8/52 = 2/13. For the second pick we have 4 "good cards" (the 4 aces if we picked a king, the 4 kings if we picked an aces) out of the 51 remaining cards, so we have a success rate of 4/51. So, the probability of picking an ace and a king, in any order, is
[tex]\dfrac{2}{13}\cdot\dfrac{4}{51}=\dfrac{8}{663}[/tex]
Using probability concepts, it is found that:
a) No. The probability of drawing a specific second card depends on the identity of the first card.
b) P(ace on 1st card and king on 2nd) = [tex]\frac{4}{663}[/tex]
c) P(king on 1st card and ace on 2nd) = [tex]\frac{4}{663}[/tex]
d) [tex]\frac{8}{663}[/tex] probability of drawing an ace and a king in either order.
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- A probability is the division of the number of desired outcomes and the number of total outcomes.
- A standard deck has 52 cards, thus, 52 is considered as the number of total outcomes.
- Of those cards, 4 are aces and 4 are kings.
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Item a:
Since the card drawn is not replaced, the trials are not independent, as the probabilities change, as for the first toss there are 52 total outcomes, and for the second there are 51. Thus, the correct option is:
No. The probability of drawing a specific second card depends on the identity of the first card.
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Item b:
- 4 of the 52 cards are aces, and then 4 of the remaining 51 will be kings. Thus:
[tex]p = \frac{4}{52} \times \frac{4}{51} = \frac{4 \times 4}{52 \times 51} = \frac{4}{13 \times 51} = \frac{4}{663}[/tex]
Thus:
P(ace on 1st card and king on 2nd) = [tex]\frac{4}{663}[/tex]
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Item c:
- 4 of the 52 cards are kings, and then 4 of the remaining 51 will be aces. Thus:
[tex]p = \frac{4}{52} \times \frac{4}{51} = \frac{4 \times 4}{52 \times 51} = \frac{4}{13 \times 51} = \frac{4}{663}[/tex]
Thus:
P(king on 1st card and ace on 2nd) = [tex]\frac{4}{663}[/tex]
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Item d:
Adding the previous two probabilities:
[tex]\frac{4}{663} + \frac{4}{663} = \frac{4 + 4}{663} = \frac{8}{663}[/tex]
[tex]\frac{8}{663}[/tex] probability of drawing an ace and a king in either order
A similar problem is given at https://brainly.com/question/24104122
