probability of drawn a ball from same bag without replacement
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Problem 1: We have 2 opaque bags, each containing 2 balls. One bag has 2 black balls and the other has a black ball and a white ball. You pick a bag at random and then pick one of the balls in that bag at random. When you look at the ball, it is black. You now pick the second ball from that same bag. What is the probability that this ball is also black?
Answer: I am little bit confuse in answer 1/2 and 3/4
Answer 1/2: There is two bags.One containing black balls and other contain 1 black and 1 white ball. If I drawn 1st ball black then the 2nd ball is also black if I chose the bag which contain 2 black ball. So to get 2nd ball also black probability is 1/2.
Answer 3/4: Total ball 4. 1st one drawn. So next one when drawn probability of getting black ball is 3/4 compare to remaining ball.
Which one is correct?
probability
edited Nov 14 at 22:33
N. F. Taussig
42.6k93254
asked Nov 14 at 21:53
RealCoder
111
add a comment |
up vote
2
down vote
favorite
Problem 1: We have 2 opaque bags, each containing 2 balls. One bag has 2 black balls and the other has a black ball and a white ball. You pick a bag at random and then pick one of the balls in that bag at random. When you look at the ball, it is black. You now pick the second ball from that same bag. What is the probability that this ball is also black?
Answer: I am little bit confuse in answer 1/2 and 3/4
Answer 1/2: There is two bags.One containing black balls and other contain 1 black and 1 white ball. If I drawn 1st ball black then the 2nd ball is also black if I chose the bag which contain 2 black ball. So to get 2nd ball also black probability is 1/2.
Answer 3/4: Total ball 4. 1st one drawn. So next one when drawn probability of getting black ball is 3/4 compare to remaining ball.
Which one is correct?
probability
edited Nov 14 at 22:33
N. F. Taussig
42.6k93254
asked Nov 14 at 21:53
RealCoder
111
Define $A$ as the event that you pick the bag with two black balls. Define $B$ as the event that the first ball you saw was black. So you want to compute $P[A|B]$, no? Do you know how to use/switch conditionings?
– Michael
Nov 14 at 22:01
What is switch conditionings?
– RealCoder
Nov 14 at 22:06
Switch conditioning. Sometimes called Baye's rule. It gives you $P[A|B]$ in terms of $P[B|A]$ and some other probabiltiies. It just applies the definition of conditional probability.
– Michael
Nov 14 at 22:07
oh okk.. I know baye's rule.
– RealCoder
Nov 14 at 22:08
Okay so you should be open to the possibility that neither of your guesses is correct. Use Baye's rule.
– Michael
Nov 14 at 22:09
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Problem 1: We have 2 opaque bags, each containing 2 balls. One bag has 2 black balls and the other has a black ball and a white ball. You pick a bag at random and then pick one of the balls in that bag at random. When you look at the ball, it is black. You now pick the second ball from that same bag. What is the probability that this ball is also black?
Answer: I am little bit confuse in answer 1/2 and 3/4
Answer 1/2: There is two bags.One containing black balls and other contain 1 black and 1 white ball. If I drawn 1st ball black then the 2nd ball is also black if I chose the bag which contain 2 black ball. So to get 2nd ball also black probability is 1/2.
Answer 3/4: Total ball 4. 1st one drawn. So next one when drawn probability of getting black ball is 3/4 compare to remaining ball.
Which one is correct?
probability
edited Nov 14 at 22:33
N. F. Taussig
42.6k93254
asked Nov 14 at 21:53
RealCoder
111
Problem 1: We have 2 opaque bags, each containing 2 balls. One bag has 2 black balls and the other has a black ball and a white ball. You pick a bag at random and then pick one of the balls in that bag at random. When you look at the ball, it is black. You now pick the second ball from that same bag. What is the probability that this ball is also black?
Answer: I am little bit confuse in answer 1/2 and 3/4
Answer 1/2: There is two bags.One containing black balls and other contain 1 black and 1 white ball. If I drawn 1st ball black then the 2nd ball is also black if I chose the bag which contain 2 black ball. So to get 2nd ball also black probability is 1/2.
Answer 3/4: Total ball 4. 1st one drawn. So next one when drawn probability of getting black ball is 3/4 compare to remaining ball.
Which one is correct?
probability
probability
edited Nov 14 at 22:33
N. F. Taussig
42.6k93254
asked Nov 14 at 21:53
RealCoder
111
edited Nov 14 at 22:33
N. F. Taussig
42.6k93254
asked Nov 14 at 21:53
RealCoder
111
edited Nov 14 at 22:33
N. F. Taussig
42.6k93254
edited Nov 14 at 22:33
N. F. Taussig
42.6k93254
edited Nov 14 at 22:33
N. F. Taussig
42.6k93254
42.6k93254
asked Nov 14 at 21:53
RealCoder
111
asked Nov 14 at 21:53
RealCoder
111
asked Nov 14 at 21:53
RealCoder
111
111
Define $A$ as the event that you pick the bag with two black balls. Define $B$ as the event that the first ball you saw was black. So you want to compute $P[A|B]$, no? Do you know how to use/switch conditionings?
– Michael
Nov 14 at 22:01
What is switch conditionings?
– RealCoder
Nov 14 at 22:06
Switch conditioning. Sometimes called Baye's rule. It gives you $P[A|B]$ in terms of $P[B|A]$ and some other probabiltiies. It just applies the definition of conditional probability.
– Michael
Nov 14 at 22:07
oh okk.. I know baye's rule.
– RealCoder
Nov 14 at 22:08
Okay so you should be open to the possibility that neither of your guesses is correct. Use Baye's rule.
– Michael
Nov 14 at 22:09
add a comment |
Define $A$ as the event that you pick the bag with two black balls. Define $B$ as the event that the first ball you saw was black. So you want to compute $P[A|B]$, no? Do you know how to use/switch conditionings?
– Michael
Nov 14 at 22:01
What is switch conditionings?
– RealCoder
Nov 14 at 22:06
Switch conditioning. Sometimes called Baye's rule. It gives you $P[A|B]$ in terms of $P[B|A]$ and some other probabiltiies. It just applies the definition of conditional probability.
– Michael
Nov 14 at 22:07
oh okk.. I know baye's rule.
– RealCoder
Nov 14 at 22:08
Okay so you should be open to the possibility that neither of your guesses is correct. Use Baye's rule.
– Michael
Nov 14 at 22:09
Define $A$ as the event that you pick the bag with two black balls. Define $B$ as the event that the first ball you saw was black. So you want to compute $P[A|B]$, no? Do you know how to use/switch conditionings?
– Michael
Nov 14 at 22:01
Define $A$ as the event that you pick the bag with two black balls. Define $B$ as the event that the first ball you saw was black. So you want to compute $P[A|B]$, no? Do you know how to use/switch conditionings?
– Michael
Nov 14 at 22:01
What is switch conditionings?
– RealCoder
Nov 14 at 22:06
What is switch conditionings?
– RealCoder
Nov 14 at 22:06
Switch conditioning. Sometimes called Baye's rule. It gives you $P[A|B]$ in terms of $P[B|A]$ and some other probabiltiies. It just applies the definition of conditional probability.
– Michael
Nov 14 at 22:07
Switch conditioning. Sometimes called Baye's rule. It gives you $P[A|B]$ in terms of $P[B|A]$ and some other probabiltiies. It just applies the definition of conditional probability.
– Michael
Nov 14 at 22:07
oh okk.. I know baye's rule.
– RealCoder
Nov 14 at 22:08
oh okk.. I know baye's rule.
– RealCoder
Nov 14 at 22:08
Okay so you should be open to the possibility that neither of your guesses is correct. Use Baye's rule.
– Michael
Nov 14 at 22:09
Okay so you should be open to the possibility that neither of your guesses is correct. Use Baye's rule.
– Michael
Nov 14 at 22:09
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
Neither is correct.
When given that you have drawn a black ball, the ball that you drew is more likely to have come from the bag with more black balls. So the answer is not $1/2$.
Now, as the second answer correctly argues, each individual ball had been equally likely to be drawn. However, on considering how many among all the black balls come from the bag with a another black ball, we find the answer is not $3/4$.
answered Nov 14 at 22:45
Graham Kemp
84.1k43378
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Neither is correct.
When given that you have drawn a black ball, the ball that you drew is more likely to have come from the bag with more black balls. So the answer is not $1/2$.
Now, as the second answer correctly argues, each individual ball had been equally likely to be drawn. However, on considering how many among all the black balls come from the bag with a another black ball, we find the answer is not $3/4$.
answered Nov 14 at 22:45
Graham Kemp
84.1k43378
add a comment |
up vote
0
down vote
Neither is correct.
When given that you have drawn a black ball, the ball that you drew is more likely to have come from the bag with more black balls. So the answer is not $1/2$.
Now, as the second answer correctly argues, each individual ball had been equally likely to be drawn. However, on considering how many among all the black balls come from the bag with a another black ball, we find the answer is not $3/4$.
answered Nov 14 at 22:45
Graham Kemp
84.1k43378
add a comment |
up vote
0
down vote
up vote
0
down vote
Neither is correct.
When given that you have drawn a black ball, the ball that you drew is more likely to have come from the bag with more black balls. So the answer is not $1/2$.
Now, as the second answer correctly argues, each individual ball had been equally likely to be drawn. However, on considering how many among all the black balls come from the bag with a another black ball, we find the answer is not $3/4$.
answered Nov 14 at 22:45
Graham Kemp
84.1k43378
Neither is correct.
When given that you have drawn a black ball, the ball that you drew is more likely to have come from the bag with more black balls. So the answer is not $1/2$.
Now, as the second answer correctly argues, each individual ball had been equally likely to be drawn. However, on considering how many among all the black balls come from the bag with a another black ball, we find the answer is not $3/4$.
answered Nov 14 at 22:45
Graham Kemp
84.1k43378
answered Nov 14 at 22:45
Graham Kemp
84.1k43378
answered Nov 14 at 22:45
Graham Kemp
84.1k43378
answered Nov 14 at 22:45
Graham Kemp
84.1k43378
84.1k43378
add a comment |
add a comment |
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Define $A$ as the event that you pick the bag with two black balls. Define $B$ as the event that the first ball you saw was black. So you want to compute $P[A|B]$, no? Do you know how to use/switch conditionings?
– Michael
Nov 14 at 22:01
What is switch conditionings?
– RealCoder
Nov 14 at 22:06
Switch conditioning. Sometimes called Baye's rule. It gives you $P[A|B]$ in terms of $P[B|A]$ and some other probabiltiies. It just applies the definition of conditional probability.
– Michael
Nov 14 at 22:07
oh okk.. I know baye's rule.
– RealCoder
Nov 14 at 22:08
Okay so you should be open to the possibility that neither of your guesses is correct. Use Baye's rule.
– Michael
Nov 14 at 22:09