Mixing
In need of tips/suggestions when to add or multiply probabilities
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I am having trouble deciding when to add or when to multiply probabilities as in the following example. I know that by constructing Probability tree diagrams we could multiply along branches and add vertically. However I could definitely use more suggestions/tips that
might help me decide when to multiply and when to add probabilities.
A jar contains $4$ black and $3$ White balls. If you reach into the jar and pick two balls simultaneously , what is the probability that one is black and the other is white ?
This is how I am solving the above : Pr(Black from the total 7 balls)=$frac{4}{7}$
Pr(White from the remaining 6 balls after choosing a Black ball) $= frac{3}{6}$
So Ans = $frac{4}{7} times frac{3}{6} = frac{2}{7}$
probability
edited Nov 18 '13 at 10:55
Greek - Area 51 Proposal
3,176769105
asked Aug 18 '12 at 22:02
MistyDMistyD
72661836
$endgroup$
add a comment |
$begingroup$
I am having trouble deciding when to add or when to multiply probabilities as in the following example. I know that by constructing Probability tree diagrams we could multiply along branches and add vertically. However I could definitely use more suggestions/tips that
might help me decide when to multiply and when to add probabilities.
A jar contains $4$ black and $3$ White balls. If you reach into the jar and pick two balls simultaneously , what is the probability that one is black and the other is white ?
This is how I am solving the above : Pr(Black from the total 7 balls)=$frac{4}{7}$
Pr(White from the remaining 6 balls after choosing a Black ball) $= frac{3}{6}$
So Ans = $frac{4}{7} times frac{3}{6} = frac{2}{7}$
probability
edited Nov 18 '13 at 10:55
Greek - Area 51 Proposal
3,176769105
asked Aug 18 '12 at 22:02
MistyDMistyD
72661836
$endgroup$
$begingroup$
It's also possible that the first one is white and the second is black.
$endgroup$
– Yuval Filmus
Aug 18 '12 at 22:17
1
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Event 1 "or" event 2 (mutually exclusive): $P_1+P_2$. Event 1 "and" event 2: $P_1times P_2$
$endgroup$
– Mike
Aug 19 '12 at 6:37
add a comment |
$begingroup$
I am having trouble deciding when to add or when to multiply probabilities as in the following example. I know that by constructing Probability tree diagrams we could multiply along branches and add vertically. However I could definitely use more suggestions/tips that
might help me decide when to multiply and when to add probabilities.
A jar contains $4$ black and $3$ White balls. If you reach into the jar and pick two balls simultaneously , what is the probability that one is black and the other is white ?
This is how I am solving the above : Pr(Black from the total 7 balls)=$frac{4}{7}$
Pr(White from the remaining 6 balls after choosing a Black ball) $= frac{3}{6}$
So Ans = $frac{4}{7} times frac{3}{6} = frac{2}{7}$
probability
edited Nov 18 '13 at 10:55
Greek - Area 51 Proposal
3,176769105
asked Aug 18 '12 at 22:02
MistyDMistyD
72661836
$endgroup$
I am having trouble deciding when to add or when to multiply probabilities as in the following example. I know that by constructing Probability tree diagrams we could multiply along branches and add vertically. However I could definitely use more suggestions/tips that
might help me decide when to multiply and when to add probabilities.
A jar contains $4$ black and $3$ White balls. If you reach into the jar and pick two balls simultaneously , what is the probability that one is black and the other is white ?
This is how I am solving the above : Pr(Black from the total 7 balls)=$frac{4}{7}$
Pr(White from the remaining 6 balls after choosing a Black ball) $= frac{3}{6}$
So Ans = $frac{4}{7} times frac{3}{6} = frac{2}{7}$
probability
probability
edited Nov 18 '13 at 10:55
Greek - Area 51 Proposal
3,176769105
asked Aug 18 '12 at 22:02
MistyDMistyD
72661836
edited Nov 18 '13 at 10:55
Greek - Area 51 Proposal
3,176769105
asked Aug 18 '12 at 22:02
MistyDMistyD
72661836
edited Nov 18 '13 at 10:55
Greek - Area 51 Proposal
3,176769105
edited Nov 18 '13 at 10:55
Greek - Area 51 Proposal
3,176769105
edited Nov 18 '13 at 10:55
Greek - Area 51 Proposal
3,176769105
3,176769105
asked Aug 18 '12 at 22:02
MistyDMistyD
72661836
asked Aug 18 '12 at 22:02
MistyDMistyD
72661836
asked Aug 18 '12 at 22:02
MistyDMistyD
72661836
72661836
$begingroup$
It's also possible that the first one is white and the second is black.
$endgroup$
– Yuval Filmus
Aug 18 '12 at 22:17
1
$begingroup$
Event 1 "or" event 2 (mutually exclusive): $P_1+P_2$. Event 1 "and" event 2: $P_1times P_2$
$endgroup$
– Mike
Aug 19 '12 at 6:37
add a comment |
$begingroup$
It's also possible that the first one is white and the second is black.
$endgroup$
– Yuval Filmus
Aug 18 '12 at 22:17
1
$begingroup$
Event 1 "or" event 2 (mutually exclusive): $P_1+P_2$. Event 1 "and" event 2: $P_1times P_2$
$endgroup$
– Mike
Aug 19 '12 at 6:37
$begingroup$
It's also possible that the first one is white and the second is black.
$endgroup$
– Yuval Filmus
Aug 18 '12 at 22:17
$begingroup$
It's also possible that the first one is white and the second is black.
$endgroup$
– Yuval Filmus
Aug 18 '12 at 22:17
1
1
$begingroup$
Event 1 "or" event 2 (mutually exclusive): $P_1+P_2$. Event 1 "and" event 2: $P_1times P_2$
$endgroup$
– Mike
Aug 19 '12 at 6:37
$begingroup$
Event 1 "or" event 2 (mutually exclusive): $P_1+P_2$. Event 1 "and" event 2: $P_1times P_2$
$endgroup$
– Mike
Aug 19 '12 at 6:37
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
You add probabilities when the events you are thinking about are alternatives [Reading score 0 goals or 1 goal or 2 goals in their match] - you are looking for "mutually exclusive" events - things which could not happen at the same time (in the same match).
You multiply probabilities when you want two or more different things to happen "at the same time" or "consecutively" [Reading score 1 and Leeds score 1 and Arsenal score 2]. The key thing here is that the events are independent - they do not affect each other, or the second does not affect the first (etc).
In your example, to get a black ball and a white ball you have two "mutually exclusive" possibilities: $1.$ white first, black second; $2.$ black first, white second.
Possibility $1$. You can choose a white first with probability $frac 3 7$ - and then you have 6 balls left, four of which are black, so $frac 4 6$ of choosing a black one. These are independent events so multiply to get $frac 3 7 times frac 4 6 = frac 2 7$.
Possibility $2$. Choosing black first then white gives $frac 4 7 times frac 3 6 = frac 2 7$.
Adding the two together gives $frac 4 7$.
It takes some skill and practice to get these right all the time. Time spent working through and understanding key examples is time well spent.
edited Feb 2 '16 at 6:02
Greek - Area 51 Proposal
3,176769105
answered Aug 18 '12 at 22:22
Mark BennetMark Bennet
81.1k982179
$endgroup$
$begingroup$
what about more than two events happening consecutively? We still multiply right?
$endgroup$
– Abcd
Nov 24 '18 at 3:46
1
$begingroup$
@Abcd Yes, but with consecutive events you have to take a little care that they are independent. Sometimes you have to break up the problem into simpler pieces so you get a collection of simple mutually exclusive events (so these you would add) and for each of the pieces you are then working with independent events (multiply). It sounds complicated, but in practice it just requires a little care.
$endgroup$
– Mark Bennet
Nov 24 '18 at 7:31
$begingroup$
Thanks, your language of explanation is very simple. Please have a look at this related question because I am getting confused by the different terminologies that everyone is using: math.stackexchange.com/questions/3011120/…
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– Abcd
Nov 24 '18 at 9:57
add a comment |
$begingroup$
Instead of picking the balls simultaneously, you can pick them one at a time, in which case you need to account for all possible orders. Your calculation only accounts for choosing Black, then White. The probability of choosing White, Black is $frac{3}{7} times frac{4}{6}$, which also comes out to $frac{2}{7}$.
Therefore, the probability of choosing White and Black in any order (that is, simultaneously), is $frac{2}{7} + frac{2}{7} = frac{4}{7}$.
answered Aug 18 '12 at 22:17
ThéophileThéophile
19.8k12946
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add a comment |
$begingroup$
Always begin by separating out the probabilities of each event, Then:
- If all the events happen (an "and question") Multiply the probabilities together.
- If only one of the events happens (an "or question") Add the probabilities together
Example:
You're buying a new car with the following options to pick from:
- 4 cylinder or 6 cylinder
- Green, Red, Blue, White
what is the probability to get a 6 cylinder AND a white car with a random selection?
As you can see on the leaf of the previous tree there are 8 different colors to choose from, so your probability if to get a car out of 8 different choices, in this case, you multiply the probability of picking a 1 car out of two engine size and 1 from 4 different colors
$$
frac{1}{2} * frac{1}{4} = frac{1}{8}
$$
answered Apr 22 '18 at 12:33
Mina GabrielMina Gabriel
1236
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Ok,let me explain.We can do this with the help of Pnc .
You have 4 black balls so choose one of them by 4c1 .choose 1 whitel ball by 3c1. U r choosing two so ur Total no of outcomes will be 7c2
So final answer-. 4c1 * 3c1/7c2...which gives 4/7...
In your method you have delebrately chosen black ball first.so u have to multiply ur and by 2! As white ball can be chosen first too
answered Jan 12 at 10:23
vishva vishva thombarevishva vishva thombare
1
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Did anyone even consider the probability of picking 2 black (BB) or 2 white (WW) the same time - probability is a function of favorable outcomes upon Total possible outcomes, we all know this; the trick is to get this right. Get the basics and you will know when to multiply and when to add. Hence, in this case it is {BB, WW, BW, WB} = Favorable outcomes!, which is a set of 4; total possible outcomes {B,B,B,B,W,W,W} = 7;
answered Apr 12 '17 at 3:49
user435901user435901
1
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add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You add probabilities when the events you are thinking about are alternatives [Reading score 0 goals or 1 goal or 2 goals in their match] - you are looking for "mutually exclusive" events - things which could not happen at the same time (in the same match).
You multiply probabilities when you want two or more different things to happen "at the same time" or "consecutively" [Reading score 1 and Leeds score 1 and Arsenal score 2]. The key thing here is that the events are independent - they do not affect each other, or the second does not affect the first (etc).
In your example, to get a black ball and a white ball you have two "mutually exclusive" possibilities: $1.$ white first, black second; $2.$ black first, white second.
Possibility $1$. You can choose a white first with probability $frac 3 7$ - and then you have 6 balls left, four of which are black, so $frac 4 6$ of choosing a black one. These are independent events so multiply to get $frac 3 7 times frac 4 6 = frac 2 7$.
Possibility $2$. Choosing black first then white gives $frac 4 7 times frac 3 6 = frac 2 7$.
Adding the two together gives $frac 4 7$.
It takes some skill and practice to get these right all the time. Time spent working through and understanding key examples is time well spent.
edited Feb 2 '16 at 6:02
Greek - Area 51 Proposal
3,176769105
answered Aug 18 '12 at 22:22
Mark BennetMark Bennet
81.1k982179
$endgroup$
$begingroup$
what about more than two events happening consecutively? We still multiply right?
$endgroup$
– Abcd
Nov 24 '18 at 3:46
1
$begingroup$
@Abcd Yes, but with consecutive events you have to take a little care that they are independent. Sometimes you have to break up the problem into simpler pieces so you get a collection of simple mutually exclusive events (so these you would add) and for each of the pieces you are then working with independent events (multiply). It sounds complicated, but in practice it just requires a little care.
$endgroup$
– Mark Bennet
Nov 24 '18 at 7:31
$begingroup$
Thanks, your language of explanation is very simple. Please have a look at this related question because I am getting confused by the different terminologies that everyone is using: math.stackexchange.com/questions/3011120/…
$endgroup$
– Abcd
Nov 24 '18 at 9:57
add a comment |
$begingroup$
You add probabilities when the events you are thinking about are alternatives [Reading score 0 goals or 1 goal or 2 goals in their match] - you are looking for "mutually exclusive" events - things which could not happen at the same time (in the same match).
You multiply probabilities when you want two or more different things to happen "at the same time" or "consecutively" [Reading score 1 and Leeds score 1 and Arsenal score 2]. The key thing here is that the events are independent - they do not affect each other, or the second does not affect the first (etc).
In your example, to get a black ball and a white ball you have two "mutually exclusive" possibilities: $1.$ white first, black second; $2.$ black first, white second.
Possibility $1$. You can choose a white first with probability $frac 3 7$ - and then you have 6 balls left, four of which are black, so $frac 4 6$ of choosing a black one. These are independent events so multiply to get $frac 3 7 times frac 4 6 = frac 2 7$.
Possibility $2$. Choosing black first then white gives $frac 4 7 times frac 3 6 = frac 2 7$.
Adding the two together gives $frac 4 7$.
It takes some skill and practice to get these right all the time. Time spent working through and understanding key examples is time well spent.
edited Feb 2 '16 at 6:02
Greek - Area 51 Proposal
3,176769105
answered Aug 18 '12 at 22:22
Mark BennetMark Bennet
81.1k982179
$endgroup$
$begingroup$
what about more than two events happening consecutively? We still multiply right?
$endgroup$
– Abcd
Nov 24 '18 at 3:46
1
$begingroup$
@Abcd Yes, but with consecutive events you have to take a little care that they are independent. Sometimes you have to break up the problem into simpler pieces so you get a collection of simple mutually exclusive events (so these you would add) and for each of the pieces you are then working with independent events (multiply). It sounds complicated, but in practice it just requires a little care.
$endgroup$
– Mark Bennet
Nov 24 '18 at 7:31
$begingroup$
Thanks, your language of explanation is very simple. Please have a look at this related question because I am getting confused by the different terminologies that everyone is using: math.stackexchange.com/questions/3011120/…
$endgroup$
– Abcd
Nov 24 '18 at 9:57
add a comment |
$begingroup$
You add probabilities when the events you are thinking about are alternatives [Reading score 0 goals or 1 goal or 2 goals in their match] - you are looking for "mutually exclusive" events - things which could not happen at the same time (in the same match).
You multiply probabilities when you want two or more different things to happen "at the same time" or "consecutively" [Reading score 1 and Leeds score 1 and Arsenal score 2]. The key thing here is that the events are independent - they do not affect each other, or the second does not affect the first (etc).
In your example, to get a black ball and a white ball you have two "mutually exclusive" possibilities: $1.$ white first, black second; $2.$ black first, white second.
Possibility $1$. You can choose a white first with probability $frac 3 7$ - and then you have 6 balls left, four of which are black, so $frac 4 6$ of choosing a black one. These are independent events so multiply to get $frac 3 7 times frac 4 6 = frac 2 7$.
Possibility $2$. Choosing black first then white gives $frac 4 7 times frac 3 6 = frac 2 7$.
Adding the two together gives $frac 4 7$.
It takes some skill and practice to get these right all the time. Time spent working through and understanding key examples is time well spent.
edited Feb 2 '16 at 6:02
Greek - Area 51 Proposal
3,176769105
answered Aug 18 '12 at 22:22
Mark BennetMark Bennet
81.1k982179
$endgroup$
You add probabilities when the events you are thinking about are alternatives [Reading score 0 goals or 1 goal or 2 goals in their match] - you are looking for "mutually exclusive" events - things which could not happen at the same time (in the same match).
You multiply probabilities when you want two or more different things to happen "at the same time" or "consecutively" [Reading score 1 and Leeds score 1 and Arsenal score 2]. The key thing here is that the events are independent - they do not affect each other, or the second does not affect the first (etc).
In your example, to get a black ball and a white ball you have two "mutually exclusive" possibilities: $1.$ white first, black second; $2.$ black first, white second.
Possibility $1$. You can choose a white first with probability $frac 3 7$ - and then you have 6 balls left, four of which are black, so $frac 4 6$ of choosing a black one. These are independent events so multiply to get $frac 3 7 times frac 4 6 = frac 2 7$.
Possibility $2$. Choosing black first then white gives $frac 4 7 times frac 3 6 = frac 2 7$.
Adding the two together gives $frac 4 7$.
It takes some skill and practice to get these right all the time. Time spent working through and understanding key examples is time well spent.
edited Feb 2 '16 at 6:02
Greek - Area 51 Proposal
3,176769105
answered Aug 18 '12 at 22:22
Mark BennetMark Bennet
81.1k982179
edited Feb 2 '16 at 6:02
Greek - Area 51 Proposal
3,176769105
edited Feb 2 '16 at 6:02
Greek - Area 51 Proposal
3,176769105
edited Feb 2 '16 at 6:02
Greek - Area 51 Proposal
3,176769105
3,176769105
answered Aug 18 '12 at 22:22
Mark BennetMark Bennet
81.1k982179
answered Aug 18 '12 at 22:22
Mark BennetMark Bennet
81.1k982179
answered Aug 18 '12 at 22:22
Mark BennetMark Bennet
81.1k982179
81.1k982179
$begingroup$
what about more than two events happening consecutively? We still multiply right?
$endgroup$
– Abcd
Nov 24 '18 at 3:46
1
$begingroup$
@Abcd Yes, but with consecutive events you have to take a little care that they are independent. Sometimes you have to break up the problem into simpler pieces so you get a collection of simple mutually exclusive events (so these you would add) and for each of the pieces you are then working with independent events (multiply). It sounds complicated, but in practice it just requires a little care.
$endgroup$
– Mark Bennet
Nov 24 '18 at 7:31
$begingroup$
Thanks, your language of explanation is very simple. Please have a look at this related question because I am getting confused by the different terminologies that everyone is using: math.stackexchange.com/questions/3011120/…
$endgroup$
– Abcd
Nov 24 '18 at 9:57
add a comment |
$begingroup$
what about more than two events happening consecutively? We still multiply right?
$endgroup$
– Abcd
Nov 24 '18 at 3:46
1
$begingroup$
@Abcd Yes, but with consecutive events you have to take a little care that they are independent. Sometimes you have to break up the problem into simpler pieces so you get a collection of simple mutually exclusive events (so these you would add) and for each of the pieces you are then working with independent events (multiply). It sounds complicated, but in practice it just requires a little care.
$endgroup$
– Mark Bennet
Nov 24 '18 at 7:31
$begingroup$
Thanks, your language of explanation is very simple. Please have a look at this related question because I am getting confused by the different terminologies that everyone is using: math.stackexchange.com/questions/3011120/…
$endgroup$
– Abcd
Nov 24 '18 at 9:57
$begingroup$
what about more than two events happening consecutively? We still multiply right?
$endgroup$
– Abcd
Nov 24 '18 at 3:46
$begingroup$
what about more than two events happening consecutively? We still multiply right?
$endgroup$
– Abcd
Nov 24 '18 at 3:46
1
1
$begingroup$
@Abcd Yes, but with consecutive events you have to take a little care that they are independent. Sometimes you have to break up the problem into simpler pieces so you get a collection of simple mutually exclusive events (so these you would add) and for each of the pieces you are then working with independent events (multiply). It sounds complicated, but in practice it just requires a little care.
$endgroup$
– Mark Bennet
Nov 24 '18 at 7:31
$begingroup$
@Abcd Yes, but with consecutive events you have to take a little care that they are independent. Sometimes you have to break up the problem into simpler pieces so you get a collection of simple mutually exclusive events (so these you would add) and for each of the pieces you are then working with independent events (multiply). It sounds complicated, but in practice it just requires a little care.
$endgroup$
– Mark Bennet
Nov 24 '18 at 7:31
$begingroup$
Thanks, your language of explanation is very simple. Please have a look at this related question because I am getting confused by the different terminologies that everyone is using: math.stackexchange.com/questions/3011120/…
$endgroup$
– Abcd
Nov 24 '18 at 9:57
$begingroup$
Thanks, your language of explanation is very simple. Please have a look at this related question because I am getting confused by the different terminologies that everyone is using: math.stackexchange.com/questions/3011120/…
$endgroup$
– Abcd
Nov 24 '18 at 9:57
add a comment |
$begingroup$
Instead of picking the balls simultaneously, you can pick them one at a time, in which case you need to account for all possible orders. Your calculation only accounts for choosing Black, then White. The probability of choosing White, Black is $frac{3}{7} times frac{4}{6}$, which also comes out to $frac{2}{7}$.
Therefore, the probability of choosing White and Black in any order (that is, simultaneously), is $frac{2}{7} + frac{2}{7} = frac{4}{7}$.
answered Aug 18 '12 at 22:17
ThéophileThéophile
19.8k12946
$endgroup$
add a comment |
$begingroup$
Instead of picking the balls simultaneously, you can pick them one at a time, in which case you need to account for all possible orders. Your calculation only accounts for choosing Black, then White. The probability of choosing White, Black is $frac{3}{7} times frac{4}{6}$, which also comes out to $frac{2}{7}$.
Therefore, the probability of choosing White and Black in any order (that is, simultaneously), is $frac{2}{7} + frac{2}{7} = frac{4}{7}$.
answered Aug 18 '12 at 22:17
ThéophileThéophile
19.8k12946
$endgroup$
add a comment |
$begingroup$
Instead of picking the balls simultaneously, you can pick them one at a time, in which case you need to account for all possible orders. Your calculation only accounts for choosing Black, then White. The probability of choosing White, Black is $frac{3}{7} times frac{4}{6}$, which also comes out to $frac{2}{7}$.
Therefore, the probability of choosing White and Black in any order (that is, simultaneously), is $frac{2}{7} + frac{2}{7} = frac{4}{7}$.
answered Aug 18 '12 at 22:17
ThéophileThéophile
19.8k12946
$endgroup$
Instead of picking the balls simultaneously, you can pick them one at a time, in which case you need to account for all possible orders. Your calculation only accounts for choosing Black, then White. The probability of choosing White, Black is $frac{3}{7} times frac{4}{6}$, which also comes out to $frac{2}{7}$.
Therefore, the probability of choosing White and Black in any order (that is, simultaneously), is $frac{2}{7} + frac{2}{7} = frac{4}{7}$.
answered Aug 18 '12 at 22:17
ThéophileThéophile
19.8k12946
answered Aug 18 '12 at 22:17
ThéophileThéophile
19.8k12946
answered Aug 18 '12 at 22:17
ThéophileThéophile
19.8k12946
answered Aug 18 '12 at 22:17
ThéophileThéophile
19.8k12946
19.8k12946
add a comment |
add a comment |
$begingroup$
Always begin by separating out the probabilities of each event, Then:
- If all the events happen (an "and question") Multiply the probabilities together.
- If only one of the events happens (an "or question") Add the probabilities together
Example:
You're buying a new car with the following options to pick from:
- 4 cylinder or 6 cylinder
- Green, Red, Blue, White
what is the probability to get a 6 cylinder AND a white car with a random selection?
As you can see on the leaf of the previous tree there are 8 different colors to choose from, so your probability if to get a car out of 8 different choices, in this case, you multiply the probability of picking a 1 car out of two engine size and 1 from 4 different colors
$$
frac{1}{2} * frac{1}{4} = frac{1}{8}
$$
answered Apr 22 '18 at 12:33
Mina GabrielMina Gabriel
1236
$endgroup$
add a comment |
$begingroup$
Always begin by separating out the probabilities of each event, Then:
- If all the events happen (an "and question") Multiply the probabilities together.
- If only one of the events happens (an "or question") Add the probabilities together
Example:
You're buying a new car with the following options to pick from:
- 4 cylinder or 6 cylinder
- Green, Red, Blue, White
what is the probability to get a 6 cylinder AND a white car with a random selection?
As you can see on the leaf of the previous tree there are 8 different colors to choose from, so your probability if to get a car out of 8 different choices, in this case, you multiply the probability of picking a 1 car out of two engine size and 1 from 4 different colors
$$
frac{1}{2} * frac{1}{4} = frac{1}{8}
$$
answered Apr 22 '18 at 12:33
Mina GabrielMina Gabriel
1236
$endgroup$
add a comment |
$begingroup$
Always begin by separating out the probabilities of each event, Then:
- If all the events happen (an "and question") Multiply the probabilities together.
- If only one of the events happens (an "or question") Add the probabilities together
Example:
You're buying a new car with the following options to pick from:
- 4 cylinder or 6 cylinder
- Green, Red, Blue, White
what is the probability to get a 6 cylinder AND a white car with a random selection?
As you can see on the leaf of the previous tree there are 8 different colors to choose from, so your probability if to get a car out of 8 different choices, in this case, you multiply the probability of picking a 1 car out of two engine size and 1 from 4 different colors
$$
frac{1}{2} * frac{1}{4} = frac{1}{8}
$$
answered Apr 22 '18 at 12:33
Mina GabrielMina Gabriel
1236
$endgroup$
Always begin by separating out the probabilities of each event, Then:
- If all the events happen (an "and question") Multiply the probabilities together.
- If only one of the events happens (an "or question") Add the probabilities together
Example:
You're buying a new car with the following options to pick from:
- 4 cylinder or 6 cylinder
- Green, Red, Blue, White
what is the probability to get a 6 cylinder AND a white car with a random selection?
As you can see on the leaf of the previous tree there are 8 different colors to choose from, so your probability if to get a car out of 8 different choices, in this case, you multiply the probability of picking a 1 car out of two engine size and 1 from 4 different colors
$$
frac{1}{2} * frac{1}{4} = frac{1}{8}
$$
answered Apr 22 '18 at 12:33
Mina GabrielMina Gabriel
1236
answered Apr 22 '18 at 12:33
Mina GabrielMina Gabriel
1236
answered Apr 22 '18 at 12:33
Mina GabrielMina Gabriel
1236
answered Apr 22 '18 at 12:33
Mina GabrielMina Gabriel
1236
1236
add a comment |
add a comment |
$begingroup$
Ok,let me explain.We can do this with the help of Pnc .
You have 4 black balls so choose one of them by 4c1 .choose 1 whitel ball by 3c1. U r choosing two so ur Total no of outcomes will be 7c2
So final answer-. 4c1 * 3c1/7c2...which gives 4/7...
In your method you have delebrately chosen black ball first.so u have to multiply ur and by 2! As white ball can be chosen first too
answered Jan 12 at 10:23
vishva vishva thombarevishva vishva thombare
1
$endgroup$
add a comment |
$begingroup$
Ok,let me explain.We can do this with the help of Pnc .
You have 4 black balls so choose one of them by 4c1 .choose 1 whitel ball by 3c1. U r choosing two so ur Total no of outcomes will be 7c2
So final answer-. 4c1 * 3c1/7c2...which gives 4/7...
In your method you have delebrately chosen black ball first.so u have to multiply ur and by 2! As white ball can be chosen first too
answered Jan 12 at 10:23
vishva vishva thombarevishva vishva thombare
1
$endgroup$
add a comment |
$begingroup$
Ok,let me explain.We can do this with the help of Pnc .
You have 4 black balls so choose one of them by 4c1 .choose 1 whitel ball by 3c1. U r choosing two so ur Total no of outcomes will be 7c2
So final answer-. 4c1 * 3c1/7c2...which gives 4/7...
In your method you have delebrately chosen black ball first.so u have to multiply ur and by 2! As white ball can be chosen first too
answered Jan 12 at 10:23
vishva vishva thombarevishva vishva thombare
1
$endgroup$
Ok,let me explain.We can do this with the help of Pnc .
You have 4 black balls so choose one of them by 4c1 .choose 1 whitel ball by 3c1. U r choosing two so ur Total no of outcomes will be 7c2
So final answer-. 4c1 * 3c1/7c2...which gives 4/7...
In your method you have delebrately chosen black ball first.so u have to multiply ur and by 2! As white ball can be chosen first too
answered Jan 12 at 10:23
vishva vishva thombarevishva vishva thombare
1
answered Jan 12 at 10:23
vishva vishva thombarevishva vishva thombare
1
answered Jan 12 at 10:23
vishva vishva thombarevishva vishva thombare
1
answered Jan 12 at 10:23
vishva vishva thombarevishva vishva thombare
1
1
add a comment |
add a comment |
$begingroup$
Did anyone even consider the probability of picking 2 black (BB) or 2 white (WW) the same time - probability is a function of favorable outcomes upon Total possible outcomes, we all know this; the trick is to get this right. Get the basics and you will know when to multiply and when to add. Hence, in this case it is {BB, WW, BW, WB} = Favorable outcomes!, which is a set of 4; total possible outcomes {B,B,B,B,W,W,W} = 7;
answered Apr 12 '17 at 3:49
user435901user435901
1
$endgroup$
add a comment |
$begingroup$
Did anyone even consider the probability of picking 2 black (BB) or 2 white (WW) the same time - probability is a function of favorable outcomes upon Total possible outcomes, we all know this; the trick is to get this right. Get the basics and you will know when to multiply and when to add. Hence, in this case it is {BB, WW, BW, WB} = Favorable outcomes!, which is a set of 4; total possible outcomes {B,B,B,B,W,W,W} = 7;
answered Apr 12 '17 at 3:49
user435901user435901
1
$endgroup$
add a comment |
$begingroup$
Did anyone even consider the probability of picking 2 black (BB) or 2 white (WW) the same time - probability is a function of favorable outcomes upon Total possible outcomes, we all know this; the trick is to get this right. Get the basics and you will know when to multiply and when to add. Hence, in this case it is {BB, WW, BW, WB} = Favorable outcomes!, which is a set of 4; total possible outcomes {B,B,B,B,W,W,W} = 7;
answered Apr 12 '17 at 3:49
user435901user435901
1
$endgroup$
Did anyone even consider the probability of picking 2 black (BB) or 2 white (WW) the same time - probability is a function of favorable outcomes upon Total possible outcomes, we all know this; the trick is to get this right. Get the basics and you will know when to multiply and when to add. Hence, in this case it is {BB, WW, BW, WB} = Favorable outcomes!, which is a set of 4; total possible outcomes {B,B,B,B,W,W,W} = 7;
answered Apr 12 '17 at 3:49
user435901user435901
1
answered Apr 12 '17 at 3:49
user435901user435901
1
answered Apr 12 '17 at 3:49
user435901user435901
1
answered Apr 12 '17 at 3:49
user435901user435901
1
1
add a comment |
add a comment |
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$begingroup$
It's also possible that the first one is white and the second is black.
$endgroup$
– Yuval Filmus
Aug 18 '12 at 22:17
1
$begingroup$
Event 1 "or" event 2 (mutually exclusive): $P_1+P_2$. Event 1 "and" event 2: $P_1times P_2$
$endgroup$
– Mike
Aug 19 '12 at 6:37