What is the probability that the second card is a king, given that the first card was a spade?
I have the following questions. Part a- the answer is 4/51 as all the jacks are still in the pack. My question is why the answer to part b not the same as all the kings are still in the pack? Instead, the answer is 1/13, why?
A pair of cards are drawn from a standard pack of 52 cards.
(a) What is the probability that the second card is a jack, given that the
first card was an ace?
(b) What is the probability that the second card is a king, given instead
that the first card was a spade?
In general, how should you approach these kind of questions? I find them difficult to visualise for harder problems..?
probability combinatorics elementary-probability
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I have the following questions. Part a- the answer is 4/51 as all the jacks are still in the pack. My question is why the answer to part b not the same as all the kings are still in the pack? Instead, the answer is 1/13, why?
A pair of cards are drawn from a standard pack of 52 cards.
(a) What is the probability that the second card is a jack, given that the
first card was an ace?
(b) What is the probability that the second card is a king, given instead
that the first card was a spade?
In general, how should you approach these kind of questions? I find them difficult to visualise for harder problems..?
probability combinatorics elementary-probability
add a comment |
I have the following questions. Part a- the answer is 4/51 as all the jacks are still in the pack. My question is why the answer to part b not the same as all the kings are still in the pack? Instead, the answer is 1/13, why?
A pair of cards are drawn from a standard pack of 52 cards.
(a) What is the probability that the second card is a jack, given that the
first card was an ace?
(b) What is the probability that the second card is a king, given instead
that the first card was a spade?
In general, how should you approach these kind of questions? I find them difficult to visualise for harder problems..?
probability combinatorics elementary-probability
I have the following questions. Part a- the answer is 4/51 as all the jacks are still in the pack. My question is why the answer to part b not the same as all the kings are still in the pack? Instead, the answer is 1/13, why?
A pair of cards are drawn from a standard pack of 52 cards.
(a) What is the probability that the second card is a jack, given that the
first card was an ace?
(b) What is the probability that the second card is a king, given instead
that the first card was a spade?
In general, how should you approach these kind of questions? I find them difficult to visualise for harder problems..?
probability combinatorics elementary-probability
probability combinatorics elementary-probability
edited Nov 21 '18 at 22:35
Asaf Karagila♦
302k32427757
302k32427757
asked Nov 21 '18 at 22:23
PumpkinpeachPumpkinpeach
628
628
add a comment |
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Hint: For part (b), either the spade was the king of spades or it wasn't.
oh how did I not think of that. Thank you
– Pumpkinpeach
Nov 21 '18 at 22:35
add a comment |
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Hint: For part (b), either the spade was the king of spades or it wasn't.
oh how did I not think of that. Thank you
– Pumpkinpeach
Nov 21 '18 at 22:35
add a comment |
Hint: For part (b), either the spade was the king of spades or it wasn't.
oh how did I not think of that. Thank you
– Pumpkinpeach
Nov 21 '18 at 22:35
add a comment |
Hint: For part (b), either the spade was the king of spades or it wasn't.
Hint: For part (b), either the spade was the king of spades or it wasn't.
answered Nov 21 '18 at 22:28
rogerlrogerl
17.4k22746
17.4k22746
oh how did I not think of that. Thank you
– Pumpkinpeach
Nov 21 '18 at 22:35
add a comment |
oh how did I not think of that. Thank you
– Pumpkinpeach
Nov 21 '18 at 22:35
oh how did I not think of that. Thank you
– Pumpkinpeach
Nov 21 '18 at 22:35
oh how did I not think of that. Thank you
– Pumpkinpeach
Nov 21 '18 at 22:35
add a comment |
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