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Probability of choosing at least one card
Probability of card distribution in bridgeProbability that a 13 card hand has at least 1 card in each suit?exactly, and at least : ProbabilityFinding the probability of choosing six numbersProbability with and without replacementProbabilty of choosing at least one ball from different boxesCould I have some help solving this conditional probability problem?Probability of making a fullhouse with a $39$ card deckProbability Card QuestionIf I split a 24 card deck evenly between three players, how do I find the probability that each player has at least one card from the hearts suit?
$begingroup$
REFER TO PART 3 (iii) OF QUESTION.
following is my solution :
Any help would be appreciated
probability
asked Mar 29 at 3:58
user122343user122343
1076
$endgroup$
add a comment |
$begingroup$
REFER TO PART 3 (iii) OF QUESTION.
following is my solution :
Any help would be appreciated
probability
asked Mar 29 at 3:58
user122343user122343
1076
$endgroup$
$begingroup$
Looks good to me.
$endgroup$
– Dbchatto67
Mar 29 at 4:39
$begingroup$
math.stackexchange.com/questions
$endgroup$
– lab bhattacharjee
Mar 29 at 4:56
add a comment |
$begingroup$
REFER TO PART 3 (iii) OF QUESTION.
following is my solution :
Any help would be appreciated
probability
asked Mar 29 at 3:58
user122343user122343
1076
$endgroup$
REFER TO PART 3 (iii) OF QUESTION.
following is my solution :
Any help would be appreciated
probability
probability
asked Mar 29 at 3:58
user122343user122343
1076
asked Mar 29 at 3:58
user122343user122343
1076
asked Mar 29 at 3:58
user122343user122343
1076
asked Mar 29 at 3:58
user122343user122343
1076
asked Mar 29 at 3:58
user122343user122343
1076
1076
$begingroup$
Looks good to me.
$endgroup$
– Dbchatto67
Mar 29 at 4:39
$begingroup$
math.stackexchange.com/questions
$endgroup$
– lab bhattacharjee
Mar 29 at 4:56
add a comment |
$begingroup$
Looks good to me.
$endgroup$
– Dbchatto67
Mar 29 at 4:39
$begingroup$
math.stackexchange.com/questions
$endgroup$
– lab bhattacharjee
Mar 29 at 4:56
$begingroup$
Looks good to me.
$endgroup$
– Dbchatto67
Mar 29 at 4:39
$begingroup$
Looks good to me.
$endgroup$
– Dbchatto67
Mar 29 at 4:39
$begingroup$
math.stackexchange.com/questions
$endgroup$
– lab bhattacharjee
Mar 29 at 4:56
$begingroup$
math.stackexchange.com/questions
$endgroup$
– lab bhattacharjee
Mar 29 at 4:56
add a comment |
1 Answer
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Yeah your solution is correct
We can solve it more easily in the following way . let $p(at least one B)$ is required probability which can also be written as
$=$ $1 - p(No B is selected)$
$=$ $1 - (6/8)*(5/7)$
$= 1 -(15/28)$
$= 13/28$
answered Mar 29 at 4:27
user1608user1608
768
$endgroup$
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
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votes
$begingroup$
Yeah your solution is correct
We can solve it more easily in the following way . let $p(at least one B)$ is required probability which can also be written as
$=$ $1 - p(No B is selected)$
$=$ $1 - (6/8)*(5/7)$
$= 1 -(15/28)$
$= 13/28$
answered Mar 29 at 4:27
user1608user1608
768
$endgroup$
add a comment |
$begingroup$
Yeah your solution is correct
We can solve it more easily in the following way . let $p(at least one B)$ is required probability which can also be written as
$=$ $1 - p(No B is selected)$
$=$ $1 - (6/8)*(5/7)$
$= 1 -(15/28)$
$= 13/28$
answered Mar 29 at 4:27
user1608user1608
768
$endgroup$
add a comment |
$begingroup$
Yeah your solution is correct
We can solve it more easily in the following way . let $p(at least one B)$ is required probability which can also be written as
$=$ $1 - p(No B is selected)$
$=$ $1 - (6/8)*(5/7)$
$= 1 -(15/28)$
$= 13/28$
answered Mar 29 at 4:27
user1608user1608
768
$endgroup$
Yeah your solution is correct
We can solve it more easily in the following way . let $p(at least one B)$ is required probability which can also be written as
$=$ $1 - p(No B is selected)$
$=$ $1 - (6/8)*(5/7)$
$= 1 -(15/28)$
$= 13/28$
answered Mar 29 at 4:27
user1608user1608
768
answered Mar 29 at 4:27
user1608user1608
768
answered Mar 29 at 4:27
user1608user1608
768
answered Mar 29 at 4:27
user1608user1608
768
768
add a comment |
add a comment |
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Looks good to me.
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– Dbchatto67
Mar 29 at 4:39
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– lab bhattacharjee
Mar 29 at 4:56