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

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5 Answers

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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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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

$$ \frac{1}{2} * \frac{1}{4} = \frac{1}{8} $$

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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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We can solve this problem using combinatorics and probability. There are 4 black balls in the jar. So, you have $\binom{4}{1}$ ways to choose one black ball. Similarly, you have $\binom{3}{1}$ ways to choose one white ball. You can choose any two balls from the jar in $\binom{7}{2}$ ways.

So, the final answer will be $$\frac{\binom{4}{1} * \binom{3}{1}}{\binom{7}{2}}$$Simplifying this, you'll get $\frac{4}{7}$.

In your method, you have deliberately chosen the black ball first. So, you'll have to multiply your answer by $2!$ as white ball can be chosen first too.

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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;

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