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# Independent events example: test taking

Have you ever taken a test and discovered you have no choice but to guess on a couple of problems? In this example problem, we are considering the probability of two independent events occurring. Created by Sal Khan and Monterey Institute for Technology and Education.

## Want to join the conversation?

• what is the difference between mutually exclusive event and independent event? •   Mutually exclusive:
The to events cannot happen at the same time. If you roll a die, you cannot get both an even and an odd.

Independent:
Knowledge that one event has occurred does not change the probability of another event occurring. For example, A=event that you go to work today, B=event that it rains today. If A occurs, you need to go to work, this has no impact on B, the probability of rain for that day.
• The easiest way I discern it is if the question includes the words "AND" vs. "OR".
"AND" means you multiply the outcomes, "OR" means you add the outcomes
( and subtract any possible overlap).
Is this too simplistic? Just curious.... • Sorry to bother but how can we, with absolute certainty, know that a given event is mutually exclusive or independent? Does Sal have a video on comparing both? • In the video Sal mentions independent events, is there such a thing as dependent events? If so, could you give a real world example. Thanks in advance. • I haven't watched the video you're talking about but I'm assuming it's probability, since we're talking about events.

Yes, for example;

If I had a huge pile of sand, and I scoop a shovel out every 10 minutes. Let's say the probability of finding a rock (making this up), is 50%. A fairly rocky sand pile but this is a made up example.

Every time I get a rock, or don't get a rock, the probability of getting one the NEXT time I shovel will be higher or lower depending on how many rocks I have left. So the second event probability, the second scoop of sand, depends on the first.

Hope I helped!

While doing the exercises at the end of this session the following question concerning flipping a coin came up: "What is the chance of heads in the first flip, heads in the second flip and heads or tails in the third flip?".
So the question was
P(HH(H OR T)) =?
This equivalent to
P( H AND H AND ( H OR T)) = P(H) * P(H) * P(H OR T) = 1/2 * 1/2 * 1 = 1/4.
So far all good.

Although I realized that P(H OR T) = 1, I tried to reason it through.
So the first question you have to pose is: "Are H and T here dependent or independent events?". I identified them as DEPENDENT, it is either H and consequently not T or the inverse, but they obviously dependent on one another. This means that I need to use the following formula:
P(H or T) = P(H) + P(T) - P(H AND T).
P(H) = 1/2
P(T) = 1/2
but what about P(H AND T)? It seemed obvious to me that it was 0, it is either H or T but never both, which gave
P(H or T) = 1/2 + 1/2 - 0 = 1 what was in line with my previous intuition.

However from watching Sall’s video’s I had developed the idea that if two events X and Y were dependent the probability of P(X AND Y) must be bigger than zero, and if they are independent P(X AND Y) = 0. But here I seemed to have a case that although X and Y are dependent the probability of X and Y = P(X AND Y) = 0.

Were did I go astray? • You went astray here:

> However from watching Sall’s video’s I had developed the idea that if two events X and Y were dependent the probability of P(X AND Y) must be bigger than zero, and if they are independent P(X AND Y) = 0

All that dependence means is that: `P(A) ≠ P(A|B)`. Using the definition of conditional probability here, we have: `P(A) ≠ P(A∩B) / P(B)`. As long as this inequality is satisfied, the two events are dependent. Note that this inequality assumes nothing about P(A∩B) except that it is a legitimate probability, i.e.: `0 ≤ P(A∩B) ≤ 1`.

So, if event A has non-zero probability, but P(A∩B)=0, then he two events must be dependent. Furthermore, since we have P(A∩B)=0 (and hence `P(A∪B) = P(A) + P(B)`), these events are called mutually exclusive.
• How do you calculate the probability of guessing the answers of a "multiple select"/"select all that apply" problem? For example, if there were 4 choices, you could choose any combination of the 4. • Have you ever heard about the "Power set"' (https://en.wikipedia.org/wiki/Power_set)? If you haven't, then here that's what that is: it's the set of all of the subsets of a certain set: let's say we have the set {1,2}: its power set would be { { }, {1}, {2}, {1,2} }. As you can see, it also includes itself and the void set. Let's now go back to the original problem: consider your "multiple select"/"select all that apply" problem as a set (I will now consider the one with 4 choices) composed by 4 elements, which represent each one of the answers: we'll therefore have set S = {A, B, C, D}. Now, let's consider a random subset G where you put all the answers you think are correct, for example {C, D}. Your question can be now converted into "How many subsets G could I take?": well, that's the amount of elements contained in the power set. The particularity of the power set is that its number of elements is equal to 2^n, where n is the number of elements of the set you're taking the power set of. Why 2^n? Well, you could say that for each one of the elements, you have 2 choices: take it into your subset G or reject it: applying this to each element will have 2^n combinations. So your answer is 2^4, which is 16.

In this case, though, you might want to consider the void set as invalid (because that would mean that no one of the answers would be correct) and your answer will be 16 - 1 - 15.

Hope this helps, if you didn't understand something of my explanation (I suck at explaining things) just reply. • Question: In Sally's sock drawer, there are two pairs of blue socks and two pairs of red socks. Sally decided to wear the first two socks she pulls from the drawer. What are the odds that her socks will match?
Can someone answer this question in simple steps, explaining how the solution is 3/7?.. • Is this where matrices begin to intersect probability then, within the context of this course? Or was that more the way he just happened to illustrate this particular problem?
(1 vote) • How can I find probability of guessing answer on problem 1 OR on problem 2?
I should add the individual probabilities? If yes, why does it work?
(1 vote) • I will assume that "OR" means one or the other or both (i.e. at least one). This is what "OR" usually means in math.

Adding the individual probabilities does not quite work, because this double-counts the probability of guessing both questions correctly. We must also subtract the probability of guessing both correctly.

From the lesson, the probability of guessing both correctly is 1/4 * 1/3 = 1/12.

So the probability of guessing question 1 OR question 2 correctly (or both) is 1/4 + 1/3 - 1/12 = 1/2.

An alternative method is to find the probability of guessing both incorrectly, then subtract from 1.

The probability is 3/4 of answering question 1 incorrectly, and 2/3 of answering question 2 incorrectly. Because the guesses are independent, the probability of guessing both incorrectly is 3/4 * 2/3 = 1/2.

So the probability of guessing question 1 correctly OR question 2 correctly (or both) is 1 - 1/2 = 1/2.