The probability of drawing at least two diamonds among three cards drawn at random with replacement
$begingroup$
I am learning Random variables and Probability distribution. I got this question some what hard! Can somebody help me solve this please.
Three cards are drawn at random successively with replacement from a well shuffled pack of cards. Getting a card of "Diamond is termed as success". What is the sum of the probabilities of which random variable '$X$' will be $X geq 2$.
Thank for your valuable time..
probability
asked Jul 28 '14 at 6:00
Ganesh VellankiGanesh Vellanki
17618
$endgroup$
add a comment |
$begingroup$
I am learning Random variables and Probability distribution. I got this question some what hard! Can somebody help me solve this please.
Three cards are drawn at random successively with replacement from a well shuffled pack of cards. Getting a card of "Diamond is termed as success". What is the sum of the probabilities of which random variable '$X$' will be $X geq 2$.
Thank for your valuable time..
probability
asked Jul 28 '14 at 6:00
Ganesh VellankiGanesh Vellanki
17618
$endgroup$
$begingroup$
What is the random variable $X$? Is it the number of diamonds? And are you asking for the probability that the number of diamonds is $ge 2$? That is a short calculation.
$endgroup$
– André Nicolas
Jul 28 '14 at 6:04
$begingroup$
Try to choose more descriptive titles in the future.
$endgroup$
– user147263
Jul 28 '14 at 6:42
add a comment |
$begingroup$
I am learning Random variables and Probability distribution. I got this question some what hard! Can somebody help me solve this please.
Three cards are drawn at random successively with replacement from a well shuffled pack of cards. Getting a card of "Diamond is termed as success". What is the sum of the probabilities of which random variable '$X$' will be $X geq 2$.
Thank for your valuable time..
probability
asked Jul 28 '14 at 6:00
Ganesh VellankiGanesh Vellanki
17618
$endgroup$
I am learning Random variables and Probability distribution. I got this question some what hard! Can somebody help me solve this please.
Three cards are drawn at random successively with replacement from a well shuffled pack of cards. Getting a card of "Diamond is termed as success". What is the sum of the probabilities of which random variable '$X$' will be $X geq 2$.
Thank for your valuable time..
probability
probability
asked Jul 28 '14 at 6:00
Ganesh VellankiGanesh Vellanki
17618
asked Jul 28 '14 at 6:00
Ganesh VellankiGanesh Vellanki
17618
edited Jul 28 '14 at 6:42
user147263
asked Jul 28 '14 at 6:00
Ganesh VellankiGanesh Vellanki
17618
asked Jul 28 '14 at 6:00
Ganesh VellankiGanesh Vellanki
17618
asked Jul 28 '14 at 6:00
Ganesh VellankiGanesh Vellanki
17618
17618
$begingroup$
What is the random variable $X$? Is it the number of diamonds? And are you asking for the probability that the number of diamonds is $ge 2$? That is a short calculation.
$endgroup$
– André Nicolas
Jul 28 '14 at 6:04
$begingroup$
Try to choose more descriptive titles in the future.
$endgroup$
– user147263
Jul 28 '14 at 6:42
add a comment |
$begingroup$
What is the random variable $X$? Is it the number of diamonds? And are you asking for the probability that the number of diamonds is $ge 2$? That is a short calculation.
$endgroup$
– André Nicolas
Jul 28 '14 at 6:04
$begingroup$
Try to choose more descriptive titles in the future.
$endgroup$
– user147263
Jul 28 '14 at 6:42
$begingroup$
What is the random variable $X$? Is it the number of diamonds? And are you asking for the probability that the number of diamonds is $ge 2$? That is a short calculation.
$endgroup$
– André Nicolas
Jul 28 '14 at 6:04
$begingroup$
What is the random variable $X$? Is it the number of diamonds? And are you asking for the probability that the number of diamonds is $ge 2$? That is a short calculation.
$endgroup$
– André Nicolas
Jul 28 '14 at 6:04
$begingroup$
Try to choose more descriptive titles in the future.
$endgroup$
– user147263
Jul 28 '14 at 6:42
$begingroup$
Try to choose more descriptive titles in the future.
$endgroup$
– user147263
Jul 28 '14 at 6:42
add a comment |
1 Answer
1
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$begingroup$
$P(X geq 2) = P(X = 2) + P(X = 3)$,
where
$P(X = k) = binom {3}{k} p^k(1-p)^{3-k}$,
with
$p = dfrac{1}{4}$ (Diamond is 1 of 4 suits)
and $k = 2, 3$ for your question.
edited Jan 19 at 6:58
PatrickT
147212
answered Jul 28 '14 at 6:07
DeepSeaDeepSea
71.2k54487
$endgroup$
$begingroup$
not understood brother..why does P(X=k) has come there...?
$endgroup$
– Ganesh Vellanki
Jul 28 '14 at 6:14
$begingroup$
Thanks I got it,, one last doubt...why is p=1/4. What made you assume that from the question...
$endgroup$
– Ganesh Vellanki
Jul 28 '14 at 6:37
1
$begingroup$
This is because the probability of getting a diamond is $frac{13}{52}=frac{1}{4}$. Sorry @8pir about interjecting.
$endgroup$
– Chinny84
Jul 28 '14 at 6:43
$begingroup$
Thank you so much....Everybody...
$endgroup$
– Ganesh Vellanki
Jul 28 '14 at 6:47
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
$P(X geq 2) = P(X = 2) + P(X = 3)$,
where
$P(X = k) = binom {3}{k} p^k(1-p)^{3-k}$,
with
$p = dfrac{1}{4}$ (Diamond is 1 of 4 suits)
and $k = 2, 3$ for your question.
edited Jan 19 at 6:58
PatrickT
147212
answered Jul 28 '14 at 6:07
DeepSeaDeepSea
71.2k54487
$endgroup$
$begingroup$
not understood brother..why does P(X=k) has come there...?
$endgroup$
– Ganesh Vellanki
Jul 28 '14 at 6:14
$begingroup$
Thanks I got it,, one last doubt...why is p=1/4. What made you assume that from the question...
$endgroup$
– Ganesh Vellanki
Jul 28 '14 at 6:37
1
$begingroup$
This is because the probability of getting a diamond is $frac{13}{52}=frac{1}{4}$. Sorry @8pir about interjecting.
$endgroup$
– Chinny84
Jul 28 '14 at 6:43
$begingroup$
Thank you so much....Everybody...
$endgroup$
– Ganesh Vellanki
Jul 28 '14 at 6:47
add a comment |
$begingroup$
$P(X geq 2) = P(X = 2) + P(X = 3)$,
where
$P(X = k) = binom {3}{k} p^k(1-p)^{3-k}$,
with
$p = dfrac{1}{4}$ (Diamond is 1 of 4 suits)
and $k = 2, 3$ for your question.
edited Jan 19 at 6:58
PatrickT
147212
answered Jul 28 '14 at 6:07
DeepSeaDeepSea
71.2k54487
$endgroup$
$begingroup$
not understood brother..why does P(X=k) has come there...?
$endgroup$
– Ganesh Vellanki
Jul 28 '14 at 6:14
$begingroup$
Thanks I got it,, one last doubt...why is p=1/4. What made you assume that from the question...
$endgroup$
– Ganesh Vellanki
Jul 28 '14 at 6:37
1
$begingroup$
This is because the probability of getting a diamond is $frac{13}{52}=frac{1}{4}$. Sorry @8pir about interjecting.
$endgroup$
– Chinny84
Jul 28 '14 at 6:43
$begingroup$
Thank you so much....Everybody...
$endgroup$
– Ganesh Vellanki
Jul 28 '14 at 6:47
add a comment |
$begingroup$
$P(X geq 2) = P(X = 2) + P(X = 3)$,
where
$P(X = k) = binom {3}{k} p^k(1-p)^{3-k}$,
with
$p = dfrac{1}{4}$ (Diamond is 1 of 4 suits)
and $k = 2, 3$ for your question.
edited Jan 19 at 6:58
PatrickT
147212
answered Jul 28 '14 at 6:07
DeepSeaDeepSea
71.2k54487
$endgroup$
$P(X geq 2) = P(X = 2) + P(X = 3)$,
where
$P(X = k) = binom {3}{k} p^k(1-p)^{3-k}$,
with
$p = dfrac{1}{4}$ (Diamond is 1 of 4 suits)
and $k = 2, 3$ for your question.
edited Jan 19 at 6:58
PatrickT
147212
answered Jul 28 '14 at 6:07
DeepSeaDeepSea
71.2k54487
edited Jan 19 at 6:58
PatrickT
147212
edited Jan 19 at 6:58
PatrickT
147212
edited Jan 19 at 6:58
PatrickT
147212
147212
answered Jul 28 '14 at 6:07
DeepSeaDeepSea
71.2k54487
answered Jul 28 '14 at 6:07
DeepSeaDeepSea
71.2k54487
answered Jul 28 '14 at 6:07
DeepSeaDeepSea
71.2k54487
71.2k54487
$begingroup$
not understood brother..why does P(X=k) has come there...?
$endgroup$
– Ganesh Vellanki
Jul 28 '14 at 6:14
$begingroup$
Thanks I got it,, one last doubt...why is p=1/4. What made you assume that from the question...
$endgroup$
– Ganesh Vellanki
Jul 28 '14 at 6:37
1
$begingroup$
This is because the probability of getting a diamond is $frac{13}{52}=frac{1}{4}$. Sorry @8pir about interjecting.
$endgroup$
– Chinny84
Jul 28 '14 at 6:43
$begingroup$
Thank you so much....Everybody...
$endgroup$
– Ganesh Vellanki
Jul 28 '14 at 6:47
add a comment |
$begingroup$
not understood brother..why does P(X=k) has come there...?
$endgroup$
– Ganesh Vellanki
Jul 28 '14 at 6:14
$begingroup$
Thanks I got it,, one last doubt...why is p=1/4. What made you assume that from the question...
$endgroup$
– Ganesh Vellanki
Jul 28 '14 at 6:37
1
$begingroup$
This is because the probability of getting a diamond is $frac{13}{52}=frac{1}{4}$. Sorry @8pir about interjecting.
$endgroup$
– Chinny84
Jul 28 '14 at 6:43
$begingroup$
Thank you so much....Everybody...
$endgroup$
– Ganesh Vellanki
Jul 28 '14 at 6:47
$begingroup$
not understood brother..why does P(X=k) has come there...?
$endgroup$
– Ganesh Vellanki
Jul 28 '14 at 6:14
$begingroup$
not understood brother..why does P(X=k) has come there...?
$endgroup$
– Ganesh Vellanki
Jul 28 '14 at 6:14
$begingroup$
Thanks I got it,, one last doubt...why is p=1/4. What made you assume that from the question...
$endgroup$
– Ganesh Vellanki
Jul 28 '14 at 6:37
$begingroup$
Thanks I got it,, one last doubt...why is p=1/4. What made you assume that from the question...
$endgroup$
– Ganesh Vellanki
Jul 28 '14 at 6:37
1
1
$begingroup$
This is because the probability of getting a diamond is $frac{13}{52}=frac{1}{4}$. Sorry @8pir about interjecting.
$endgroup$
– Chinny84
Jul 28 '14 at 6:43
$begingroup$
This is because the probability of getting a diamond is $frac{13}{52}=frac{1}{4}$. Sorry @8pir about interjecting.
$endgroup$
– Chinny84
Jul 28 '14 at 6:43
$begingroup$
Thank you so much....Everybody...
$endgroup$
– Ganesh Vellanki
Jul 28 '14 at 6:47
$begingroup$
Thank you so much....Everybody...
$endgroup$
– Ganesh Vellanki
Jul 28 '14 at 6:47
add a comment |
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$begingroup$
What is the random variable $X$? Is it the number of diamonds? And are you asking for the probability that the number of diamonds is $ge 2$? That is a short calculation.
$endgroup$
– André Nicolas
Jul 28 '14 at 6:04
$begingroup$
Try to choose more descriptive titles in the future.
$endgroup$
– user147263
Jul 28 '14 at 6:42