probability of drawing three hearts union/intersecting drawing at most 1 king
I have a question that is asking the probability of the intersection and union of two cases. The question assumes you are drawing 5 cards from a deck of 52.
A: Drawing at least 3 hearts
B: Drawing at most 1 King
I have solved the two as
$P(A)frac{binom{13}{4}binom{39}{1}}{binom{52}{5}}+frac{binom{13}{3}binom{39}{2}}{binom{52}{5}}+frac{binom{13}{5}}{binom{52}{5}}=9.28%$ and
$P(B)frac{binom{4}{1}binom{48}{4}}{binom{52}{5}}+frac{binom{48}{5}}{binom{52}{5}}=95.83%$
So now I cannot quite figure out how to $P(Acap B)$ and $P(Acup B)$. For $P(Acap B)$ I am not sure what $P(B|A)$ would be. I know there are two cases:
- Three hearts and NO kings
- Three hearts and ONE king
If anyone can provide some pointers it would be very appreciated.
probability combinatorics
asked Nov 27 at 2:24
Terry
62
add a comment |
I have a question that is asking the probability of the intersection and union of two cases. The question assumes you are drawing 5 cards from a deck of 52.
A: Drawing at least 3 hearts
B: Drawing at most 1 King
I have solved the two as
$P(A)frac{binom{13}{4}binom{39}{1}}{binom{52}{5}}+frac{binom{13}{3}binom{39}{2}}{binom{52}{5}}+frac{binom{13}{5}}{binom{52}{5}}=9.28%$ and
$P(B)frac{binom{4}{1}binom{48}{4}}{binom{52}{5}}+frac{binom{48}{5}}{binom{52}{5}}=95.83%$
So now I cannot quite figure out how to $P(Acap B)$ and $P(Acup B)$. For $P(Acap B)$ I am not sure what $P(B|A)$ would be. I know there are two cases:
- Three hearts and NO kings
- Three hearts and ONE king
If anyone can provide some pointers it would be very appreciated.
probability combinatorics
asked Nov 27 at 2:24
Terry
62
Actually you have made a great start in calculating $P(A), P(B)$. In the process, you have actually partitioned $A$ into $3$ events as $3, 4, 5$ hearts and $B$ into $2$ events as $0, 1$ kings. So when consider the intersection $A cap B$, you naturally have $3 times 2 = 6$ partitions to be considered, and you have named the first $2$. When you are counting, I recommend you to partition the deck into $4$ groups, namely the heart king, the remaining $12$ hearts, the remaining $3$ kings, and the remaining $52 - 1 - 12 - 3 = 36$ cards. For the union, you just apply inclusion-exclusion principle.
– BGM
Nov 27 at 6:05
add a comment |
I have a question that is asking the probability of the intersection and union of two cases. The question assumes you are drawing 5 cards from a deck of 52.
A: Drawing at least 3 hearts
B: Drawing at most 1 King
I have solved the two as
$P(A)frac{binom{13}{4}binom{39}{1}}{binom{52}{5}}+frac{binom{13}{3}binom{39}{2}}{binom{52}{5}}+frac{binom{13}{5}}{binom{52}{5}}=9.28%$ and
$P(B)frac{binom{4}{1}binom{48}{4}}{binom{52}{5}}+frac{binom{48}{5}}{binom{52}{5}}=95.83%$
So now I cannot quite figure out how to $P(Acap B)$ and $P(Acup B)$. For $P(Acap B)$ I am not sure what $P(B|A)$ would be. I know there are two cases:
- Three hearts and NO kings
- Three hearts and ONE king
If anyone can provide some pointers it would be very appreciated.
probability combinatorics
asked Nov 27 at 2:24
Terry
62
I have a question that is asking the probability of the intersection and union of two cases. The question assumes you are drawing 5 cards from a deck of 52.
A: Drawing at least 3 hearts
B: Drawing at most 1 King
I have solved the two as
$P(A)frac{binom{13}{4}binom{39}{1}}{binom{52}{5}}+frac{binom{13}{3}binom{39}{2}}{binom{52}{5}}+frac{binom{13}{5}}{binom{52}{5}}=9.28%$ and
$P(B)frac{binom{4}{1}binom{48}{4}}{binom{52}{5}}+frac{binom{48}{5}}{binom{52}{5}}=95.83%$
So now I cannot quite figure out how to $P(Acap B)$ and $P(Acup B)$. For $P(Acap B)$ I am not sure what $P(B|A)$ would be. I know there are two cases:
- Three hearts and NO kings
- Three hearts and ONE king
If anyone can provide some pointers it would be very appreciated.
probability combinatorics
probability combinatorics
asked Nov 27 at 2:24
Terry
62
asked Nov 27 at 2:24
Terry
62
asked Nov 27 at 2:24
Terry
62
asked Nov 27 at 2:24
Terry
62
asked Nov 27 at 2:24
Terry
62
62
Actually you have made a great start in calculating $P(A), P(B)$. In the process, you have actually partitioned $A$ into $3$ events as $3, 4, 5$ hearts and $B$ into $2$ events as $0, 1$ kings. So when consider the intersection $A cap B$, you naturally have $3 times 2 = 6$ partitions to be considered, and you have named the first $2$. When you are counting, I recommend you to partition the deck into $4$ groups, namely the heart king, the remaining $12$ hearts, the remaining $3$ kings, and the remaining $52 - 1 - 12 - 3 = 36$ cards. For the union, you just apply inclusion-exclusion principle.
– BGM
Nov 27 at 6:05
add a comment |
Actually you have made a great start in calculating $P(A), P(B)$. In the process, you have actually partitioned $A$ into $3$ events as $3, 4, 5$ hearts and $B$ into $2$ events as $0, 1$ kings. So when consider the intersection $A cap B$, you naturally have $3 times 2 = 6$ partitions to be considered, and you have named the first $2$. When you are counting, I recommend you to partition the deck into $4$ groups, namely the heart king, the remaining $12$ hearts, the remaining $3$ kings, and the remaining $52 - 1 - 12 - 3 = 36$ cards. For the union, you just apply inclusion-exclusion principle.
– BGM
Nov 27 at 6:05
Actually you have made a great start in calculating $P(A), P(B)$. In the process, you have actually partitioned $A$ into $3$ events as $3, 4, 5$ hearts and $B$ into $2$ events as $0, 1$ kings. So when consider the intersection $A cap B$, you naturally have $3 times 2 = 6$ partitions to be considered, and you have named the first $2$. When you are counting, I recommend you to partition the deck into $4$ groups, namely the heart king, the remaining $12$ hearts, the remaining $3$ kings, and the remaining $52 - 1 - 12 - 3 = 36$ cards. For the union, you just apply inclusion-exclusion principle.
– BGM
Nov 27 at 6:05
Actually you have made a great start in calculating $P(A), P(B)$. In the process, you have actually partitioned $A$ into $3$ events as $3, 4, 5$ hearts and $B$ into $2$ events as $0, 1$ kings. So when consider the intersection $A cap B$, you naturally have $3 times 2 = 6$ partitions to be considered, and you have named the first $2$. When you are counting, I recommend you to partition the deck into $4$ groups, namely the heart king, the remaining $12$ hearts, the remaining $3$ kings, and the remaining $52 - 1 - 12 - 3 = 36$ cards. For the union, you just apply inclusion-exclusion principle.
– BGM
Nov 27 at 6:05
add a comment |
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Actually you have made a great start in calculating $P(A), P(B)$. In the process, you have actually partitioned $A$ into $3$ events as $3, 4, 5$ hearts and $B$ into $2$ events as $0, 1$ kings. So when consider the intersection $A cap B$, you naturally have $3 times 2 = 6$ partitions to be considered, and you have named the first $2$. When you are counting, I recommend you to partition the deck into $4$ groups, namely the heart king, the remaining $12$ hearts, the remaining $3$ kings, and the remaining $52 - 1 - 12 - 3 = 36$ cards. For the union, you just apply inclusion-exclusion principle.
– BGM
Nov 27 at 6:05