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## Precalculus (2018 edition)

### Course: Precalculus (2018 edition)>Unit 6

Lesson 5: Dependent events

# Dependent probability example

It's important to practice these probability problems as they get more complex eventually. Take a stab on this one...with our help, of course. Created by Sal Khan.

## Want to join the conversation?

• What if you pick a new coin for every flip instead of keeping it? It seems to me it would be mindbogglingly difficult to calculate it, is this true? •  No, it is not difficult.

If you were to pick a new coin for every flip, then you would be evaluating "what are the chances of getting a coin out of the bag, flipping it, getting a Heads, and doing it all over again four times in a row?". You would first need to determine what the chances are for getting a coin and getting a Heads, and then multiplying this value by itself 4 times.

The chances for getting a coin and getting a Heads, it would be the addition of the chances of getting a Fair coin and getting a Heads, plus the chances of getting an Unfair coin and getting a Heads. So, (1/4)*0.5 + (3/4)*0.55 = 53.75%. This is the probability of getting a coin, any coin, and getting a Heads.

To determine the chances of getting four of these situations in a row, simply multiply 0.5375 times itself four times. So, (0.5375)^4 = 8.35%
• Why do you add the prob of the unfair coin to the prob of the fair coin? why not multiply their probabilities? •  When the probability is about A AND B, then you multiply. For example, to find the probability of getting fair coin AND 4 heads you need to multiply.
When the probability is A OR B, you add. To find the probability of getting fair coin OR unfair coin, you added their probabilities.
• Why is the following calculation incorrect?
And if so, what is the difference in the situation with my calculation and the one on the video?
If I calculate the odds of gettin one heads in one throw -> (3/4*55% + 1/4*50%)) = 53,75%
Now the odds of getting 4 in a row would be 53,75^4=8,346...
And the result in the video was 8,425...
EDIT: I guess this was answered already by Jean Rambo... But the interesting point is that there is only 0,079% advantage in picking a coin and using it 4 times than picking a coin 4 times and then flipping it... • I think the difference can be better understood by setting both scenarios in this way:

When the probability gives you 0.08425 is under the circumstance where you are asked to pick any coin from the bag and then flip it 4 times, hoping to get heads (as the problem states).

When the probability gives you 0.08347 is like you were asked to pick any coin, flip it just one time hoping to get heads, put it back in the bag, and repeat this whole process 4 times.

Now we can conclude that the slight difference in these probabilities is due to the persistent possibility (in the 2nd case) of getting a fair coin in each of the 4 flips (which offers a lower probability of getting heads than an unfair one), while in the first case that possibility just occurs once.
• at , why are the probabilities added and not multiplied? when should probabilities be added and when multiplied??
(1 vote) • Probabilities are generally multiplied when something is done over and over in a row ( ex. flipping a coin 4 times and getting heads 4 times in a row is .5^4 or .5 * .5 * .5 *.5 ).
Probabilities are generally added together when you are trying to find a total probability between two independent probabilities on a probability tree ( ex. the whole video shows a probability chart. )
• sorry, i have a question about the exercise,the question is if haven't already been hit,
captain emily have 1/2 chance of hitting pirate ashley,if captain emily is hit,then she
always miss,if haven't already been hit,pirate ashley have 1/7chance of hitting captain
emily,is she is hit,she always miss,the question is :if emily fires first,what is the probability of emily hits,ashley missed.my understanding is emily has 1/2chance times (1-1/7)=6/7,i know it's wrong,i checked the hints.but the hints says the pirate is already hit,so 1/2*(1-emily missed)=1/2*(1-0)=1/2,but it didn't state in the question that pirate has • so if the probabilities are A and B you have to multiply, and if they are A or B you have to add? just trying to make sure I understand • Still unclear about this: This is called dependent probability, but we're multiplying prob's together as if they were independent.
How does this fit with what he was saying earlier about you can multiply prob's together if they are independent? • If they are independent, then P(A and B) = P(A)^P(B) which is just the probability of A times the probability of B.
If they are dependent, then P(A and B) = P(A)*P(B|A) which is the probability of A times the probability of "B happening if A has occurred," which is different than the "Probability of B if A has not occurred."
• what i don't understand is why he doesn't lay out the procedure for finding P(A&B) from P(A|B) while incorporating P(B). this is the layout of one of my quiz questions. can anyone explain this? • Can't this be done like that,
For 4 coins avg probability of getting heads is (.55+.55+.55+.5)/4 which is = 0.5375 now for each flip this is the avg probability of each coin of getting heads thus multiply it 4 times itself for each flip event which is = 0.08346682129?
There is a difference between my answer and Sal's of about minor 0.93% only but i feel this method is right one, isn't it? • 2.02 Why is it 0.55 TIMES o.55 instead of 4 times 0.55? I don't see the reason to the repeated multiplication.
(1 vote) • 4·0.55 = 2.2, that would mean that there's a 220% probability of this event occurring, and that doesn't make much sense since the probability can't exceed a 100%.

Imagine if you were flipping not one, but some huge number of coins, let's say, a million of them. You would expect, that given 55% chance that each coin will land heads, after flipping all of them, you would end up with roughly 550 thousands heads, or 1,000,000·0.55. If you then take the coins that landed heads, and flip them all a second time, out of 550 thousands of them, roughly 302,500 or 550,000·0.55 will land heads once again. So the expected amount of coins out of million that will land heads on both flips is 1,000,000·0.55·0.55 or 1,000,000·0.55². The expected amount of coins that will land heads on each of three flips would be about 1,000,000·0.55³ and so on.

With that in mind, if you pick one of the million initial coins and ask what is the probability that this would be one of the coins that lands heads three times in a row, then that would be: (1,000,000·0.55³)/1,000,000 = 0.55³