# In need of tips/suggestions when to add or multiply probabilities

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 = $\frac{4}{7} \times \frac{3}{6} = \frac{2}{7}$

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It's also possible that the first one is white and the second is black. – Yuval Filmus Aug 18 '12 at 22:17
Event 1 "or" event 2 (mutually exclusive): $P_1+P_2$. Event 1 "and" event 2: $P_1\times P_2$ – Mike Aug 19 '12 at 6:37

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.

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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 $\frac{2}{7} + \frac{2}{7} = \frac{4}{7}$.

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I'm shocked at the terrible answers you were given, mostly how they completely ignore the word 'simultaneously'.

A good way is trial and error. The probabilities are 4/7 white and 3/7 for black, because you're picking them at the same time. If you reduced one to /6, which one would you choose? It completely screws up the equation.

So it is 4/7 and 3/7. If you add them, you get 7/7 or 1, which is 'certain'. You know it can't be this, because it's not certain you'll pick one of each colour. So it must be multiply: 4/7 x 3/7 = 12/49.

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-1: Trial and error is definitely not "a good way", as illustrated by the fact that you got the wrong answer. If you really want to take the word "simultaneously" at face value, the probability is the ratio of the number of possible pairs with one black and one white ball to the total number of possible pairs. This equals $(4\times 3)/\binom72=4/7$, validating the other answers. – Rahul Feb 27 '13 at 20:43
The answer is 12/49, Rahul. Your answer is not vindicated by the other answers as they change the question before answering it. "Instead of picking the balls simultaneously, you can pick them one at a time, in which case...", for example. I feel sorry for the guy with the question, s/he is being given bad info. – user64262 Mar 20 '13 at 21:35
"the probability is the ratio of the number of possible pairs with one black and one white ball to the total number of possible pairs. This equals $(4\times3)/\binom72=4/7$" – Rahul Mar 20 '13 at 22:13
"The probabilities are 4/7 white and 3/7 for black, because you're picking them at the same time. If you reduced one to /6, which one would you choose? It completely screws up the equation." – user64262 Mar 25 '13 at 22:35
Ah, hang on, I've read it properly. What you do here is multiply the number of one type of ball you have to pick by the second type of ball you have to pick. In this case 4 x 3, because you need one of white and black, of which there are 4 and 3 respectively. Then you times the number of balls by the number of balls minus 1, then divide by the amount of balls you're pcking out. In this case: 7 x 6 divided by 2 = 21. You put the answer of your first equation over the answer of the second: 12/21, which is simplified to 4/7 Hey, at least I was the only one to actually explain it. – user64262 Mar 25 '13 at 22:46