Mixing
Consider of drawing one card from a deck of $52$. Prove that the events of a spade being drawn and an ace...
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Consider of drawing one card from a deck of $52$. Prove that the events of a spade being drawn and an ace being drawn are independent events.
Let $A$ be the event that a spade is drawn and let $B$ be the event that an ace is drawn.
Then, $text P(A) = 4/52 = 1/13$ and $text P(B) = 4/52 = 1/13$.
How can I calculate $text P(Acap B)$? And how can I prove that these events are independent since the question specifically asked to prove that they are independent?
probability independence
edited Jan 2 at 16:36
TheSimpliFire
12.6k62360
asked Jan 2 at 16:30
Sakir InteserSakir Inteser
31
$endgroup$
add a comment |
$begingroup$
Consider of drawing one card from a deck of $52$. Prove that the events of a spade being drawn and an ace being drawn are independent events.
Let $A$ be the event that a spade is drawn and let $B$ be the event that an ace is drawn.
Then, $text P(A) = 4/52 = 1/13$ and $text P(B) = 4/52 = 1/13$.
How can I calculate $text P(Acap B)$? And how can I prove that these events are independent since the question specifically asked to prove that they are independent?
probability independence
edited Jan 2 at 16:36
TheSimpliFire
12.6k62360
asked Jan 2 at 16:30
Sakir InteserSakir Inteser
31
$endgroup$
$begingroup$
What card did you get in $A cap B$? What is the probability that you drew that card? What is the definition of independence?
$endgroup$
– Ethan Bolker
Jan 2 at 16:32
$begingroup$
$P(Acap B)$ is the probability that the card drawn is both an ace and a spade.
$endgroup$
– Lord Shark the Unknown
Jan 2 at 16:32
$begingroup$
Independence is determined by $text P(Acap B)=text P(A)timestext P(B)$
$endgroup$
– TheSimpliFire
Jan 2 at 16:33
2
$begingroup$
In a deck of 52 cards there should be more than four spades be present ...
$endgroup$
– Michael Hoppe
Jan 2 at 16:33
$begingroup$
Is there really a $1/13$ probability of drawing a spade?
$endgroup$
– Lord Shark the Unknown
Jan 2 at 16:33
add a comment |
$begingroup$
Consider of drawing one card from a deck of $52$. Prove that the events of a spade being drawn and an ace being drawn are independent events.
Let $A$ be the event that a spade is drawn and let $B$ be the event that an ace is drawn.
Then, $text P(A) = 4/52 = 1/13$ and $text P(B) = 4/52 = 1/13$.
How can I calculate $text P(Acap B)$? And how can I prove that these events are independent since the question specifically asked to prove that they are independent?
probability independence
edited Jan 2 at 16:36
TheSimpliFire
12.6k62360
asked Jan 2 at 16:30
Sakir InteserSakir Inteser
31
$endgroup$
Consider of drawing one card from a deck of $52$. Prove that the events of a spade being drawn and an ace being drawn are independent events.
Let $A$ be the event that a spade is drawn and let $B$ be the event that an ace is drawn.
Then, $text P(A) = 4/52 = 1/13$ and $text P(B) = 4/52 = 1/13$.
How can I calculate $text P(Acap B)$? And how can I prove that these events are independent since the question specifically asked to prove that they are independent?
probability independence
probability independence
edited Jan 2 at 16:36
TheSimpliFire
12.6k62360
asked Jan 2 at 16:30
Sakir InteserSakir Inteser
31
edited Jan 2 at 16:36
TheSimpliFire
12.6k62360
asked Jan 2 at 16:30
Sakir InteserSakir Inteser
31
edited Jan 2 at 16:36
TheSimpliFire
12.6k62360
edited Jan 2 at 16:36
TheSimpliFire
12.6k62360
edited Jan 2 at 16:36
TheSimpliFire
12.6k62360
12.6k62360
asked Jan 2 at 16:30
Sakir InteserSakir Inteser
31
asked Jan 2 at 16:30
Sakir InteserSakir Inteser
31
asked Jan 2 at 16:30
Sakir InteserSakir Inteser
31
31
$begingroup$
What card did you get in $A cap B$? What is the probability that you drew that card? What is the definition of independence?
$endgroup$
– Ethan Bolker
Jan 2 at 16:32
$begingroup$
$P(Acap B)$ is the probability that the card drawn is both an ace and a spade.
$endgroup$
– Lord Shark the Unknown
Jan 2 at 16:32
$begingroup$
Independence is determined by $text P(Acap B)=text P(A)timestext P(B)$
$endgroup$
– TheSimpliFire
Jan 2 at 16:33
2
$begingroup$
In a deck of 52 cards there should be more than four spades be present ...
$endgroup$
– Michael Hoppe
Jan 2 at 16:33
$begingroup$
Is there really a $1/13$ probability of drawing a spade?
$endgroup$
– Lord Shark the Unknown
Jan 2 at 16:33
add a comment |
$begingroup$
What card did you get in $A cap B$? What is the probability that you drew that card? What is the definition of independence?
$endgroup$
– Ethan Bolker
Jan 2 at 16:32
$begingroup$
$P(Acap B)$ is the probability that the card drawn is both an ace and a spade.
$endgroup$
– Lord Shark the Unknown
Jan 2 at 16:32
$begingroup$
Independence is determined by $text P(Acap B)=text P(A)timestext P(B)$
$endgroup$
– TheSimpliFire
Jan 2 at 16:33
2
$begingroup$
In a deck of 52 cards there should be more than four spades be present ...
$endgroup$
– Michael Hoppe
Jan 2 at 16:33
$begingroup$
Is there really a $1/13$ probability of drawing a spade?
$endgroup$
– Lord Shark the Unknown
Jan 2 at 16:33
$begingroup$
What card did you get in $A cap B$? What is the probability that you drew that card? What is the definition of independence?
$endgroup$
– Ethan Bolker
Jan 2 at 16:32
$begingroup$
What card did you get in $A cap B$? What is the probability that you drew that card? What is the definition of independence?
$endgroup$
– Ethan Bolker
Jan 2 at 16:32
$begingroup$
$P(Acap B)$ is the probability that the card drawn is both an ace and a spade.
$endgroup$
– Lord Shark the Unknown
Jan 2 at 16:32
$begingroup$
$P(Acap B)$ is the probability that the card drawn is both an ace and a spade.
$endgroup$
– Lord Shark the Unknown
Jan 2 at 16:32
$begingroup$
Independence is determined by $text P(Acap B)=text P(A)timestext P(B)$
$endgroup$
– TheSimpliFire
Jan 2 at 16:33
$begingroup$
Independence is determined by $text P(Acap B)=text P(A)timestext P(B)$
$endgroup$
– TheSimpliFire
Jan 2 at 16:33
2
2
$begingroup$
In a deck of 52 cards there should be more than four spades be present ...
$endgroup$
– Michael Hoppe
Jan 2 at 16:33
$begingroup$
In a deck of 52 cards there should be more than four spades be present ...
$endgroup$
– Michael Hoppe
Jan 2 at 16:33
$begingroup$
Is there really a $1/13$ probability of drawing a spade?
$endgroup$
– Lord Shark the Unknown
Jan 2 at 16:33
$begingroup$
Is there really a $1/13$ probability of drawing a spade?
$endgroup$
– Lord Shark the Unknown
Jan 2 at 16:33
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You didn't compute the probability of event $A$ correctly. There are $13$ spades in a standard deck. So $P(A)=13/52=1/4$. Note that $Acap B$ corresponds to drawing the ace of spades and hence
$$
frac{1}{52}=P(Acap B)=frac{1}{4}timesfrac{1}{13}=P(A)P(B)
$$
answered Jan 2 at 16:34
Foobaz JohnFoobaz John
21.6k41352
$endgroup$
$begingroup$
Thank you so very much!!!
$endgroup$
– Sakir Inteser
Jan 2 at 16:50
add a comment |
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1 Answer
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active
oldest
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1 Answer
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$begingroup$
You didn't compute the probability of event $A$ correctly. There are $13$ spades in a standard deck. So $P(A)=13/52=1/4$. Note that $Acap B$ corresponds to drawing the ace of spades and hence
$$
frac{1}{52}=P(Acap B)=frac{1}{4}timesfrac{1}{13}=P(A)P(B)
$$
answered Jan 2 at 16:34
Foobaz JohnFoobaz John
21.6k41352
$endgroup$
$begingroup$
Thank you so very much!!!
$endgroup$
– Sakir Inteser
Jan 2 at 16:50
add a comment |
$begingroup$
You didn't compute the probability of event $A$ correctly. There are $13$ spades in a standard deck. So $P(A)=13/52=1/4$. Note that $Acap B$ corresponds to drawing the ace of spades and hence
$$
frac{1}{52}=P(Acap B)=frac{1}{4}timesfrac{1}{13}=P(A)P(B)
$$
answered Jan 2 at 16:34
Foobaz JohnFoobaz John
21.6k41352
$endgroup$
$begingroup$
Thank you so very much!!!
$endgroup$
– Sakir Inteser
Jan 2 at 16:50
add a comment |
$begingroup$
You didn't compute the probability of event $A$ correctly. There are $13$ spades in a standard deck. So $P(A)=13/52=1/4$. Note that $Acap B$ corresponds to drawing the ace of spades and hence
$$
frac{1}{52}=P(Acap B)=frac{1}{4}timesfrac{1}{13}=P(A)P(B)
$$
answered Jan 2 at 16:34
Foobaz JohnFoobaz John
21.6k41352
$endgroup$
You didn't compute the probability of event $A$ correctly. There are $13$ spades in a standard deck. So $P(A)=13/52=1/4$. Note that $Acap B$ corresponds to drawing the ace of spades and hence
$$
frac{1}{52}=P(Acap B)=frac{1}{4}timesfrac{1}{13}=P(A)P(B)
$$
answered Jan 2 at 16:34
Foobaz JohnFoobaz John
21.6k41352
answered Jan 2 at 16:34
Foobaz JohnFoobaz John
21.6k41352
answered Jan 2 at 16:34
Foobaz JohnFoobaz John
21.6k41352
answered Jan 2 at 16:34
Foobaz JohnFoobaz John
21.6k41352
21.6k41352
$begingroup$
Thank you so very much!!!
$endgroup$
– Sakir Inteser
Jan 2 at 16:50
add a comment |
$begingroup$
Thank you so very much!!!
$endgroup$
– Sakir Inteser
Jan 2 at 16:50
$begingroup$
Thank you so very much!!!
$endgroup$
– Sakir Inteser
Jan 2 at 16:50
$begingroup$
Thank you so very much!!!
$endgroup$
– Sakir Inteser
Jan 2 at 16:50
add a comment |
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$begingroup$
What card did you get in $A cap B$? What is the probability that you drew that card? What is the definition of independence?
$endgroup$
– Ethan Bolker
Jan 2 at 16:32
$begingroup$
$P(Acap B)$ is the probability that the card drawn is both an ace and a spade.
$endgroup$
– Lord Shark the Unknown
Jan 2 at 16:32
$begingroup$
Independence is determined by $text P(Acap B)=text P(A)timestext P(B)$
$endgroup$
– TheSimpliFire
Jan 2 at 16:33
2
$begingroup$
In a deck of 52 cards there should be more than four spades be present ...
$endgroup$
– Michael Hoppe
Jan 2 at 16:33
$begingroup$
Is there really a $1/13$ probability of drawing a spade?
$endgroup$
– Lord Shark the Unknown
Jan 2 at 16:33