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## Statistics and probability

### Unit 7: Lesson 8

Multiplication rule for dependent events

# Dependent probability: coins

We're thinking about how the probability of an event can be dependent on another event occuring in this example problem. Created by Sal Khan.

## Want to join the conversation?

• why do we multiply 5/8 by 0.25? i mean why do we multiply? why cant we add? whats the significance of multiplication? and why do we add the probabilities at last?
• Multiplying 2 probabilities together calculates the probability of them both occurring. We multiply 5/8 by 0.25 to determine the probability of drawing a fair coin and then flipping 2 heads. We then went on to multiply 3/8 by 0.36 to determine the probability of drawing an unfair coin and flipping 2 heads.

Adding 2 probabilities together calculates the probability that either outcome will occur. We added 0.15625 and 0.135 (the outcomes from our multiplication problems above) together because either of those probabilities result in 2 heads being flipped. The problem never specifies what type of coin is drawn, so we add them together to find the total probability of flipping 2 heads.

Hope this helps!
• Am I safe to assume that we are replacing the coins after we've taken them out of the bag? Sal didn't take into account that the probability of drawing a fair or unfair coin would change if the first coin wasn't replaced. How would drawing a coin without replacement change the way we approach this question?
• No; we’re taking exactly one coin and flipping that same coin twice.
• if events are dependent why didn't sal use the conditional formula
p(a|b)=p(a∩b)/p(b),and why did he multiplying the possibility togther?
• Would you just add up a 3-branch probability tree if you were choosing 2 coins (thus allowing the possibility of both 1 fair and 1 unfair) to flip instead of just 1?
• If picking coins were independent events, you could use a 3-branch tree (2 fair, 2 unfair, one of each). Since in this problem the probability of picking your second coin depends on the first coin you picked, you use a 2 branch tree to pick your first coin, and then on each branch is another 2 branch tree.

This leads to 4 possible outcomes: (Fair, Fair), (Fair, Unfair), (Unfair, Fair), (Unfair, Unfair). The probability of choosing unfair first is 3/8. After an unfair is chosen, there are 5 fairs and two unfairs left in the bag, and so the probability of choosing (Unfair, Fair) is 3/8*5/7 = 15/56. On the other hand, the probability of getting (Fair, Unfair) is p(1st Fair) * p(2nd Unfair | 1st Fair) = 5/8 * 3/7, because after picking a fair coin, the bag has 4 fair and 3 unfair left.
(1 vote)
• I don't know if maybe they mention probability symbols in other video but could anyone tell me what's the symbol for "either A or B but not both"?
(1 vote)
• What you refer to is called XOR - "Exclusive Or". It does not have a standard symbol, but is often represented by a circle with a diagonal cross in it, or a v with a bar below it.
• So, if we have independent events, we add the probabilities and if we have dependent events, we multiply the probabilities?
(1 vote)
• No, but you're close....

Independent events are two events where the first one doesn't affect the other one, and if you are doing two events in a row you multiply them. Example: If you flip a coin you can get two results, your probability of getting one side is 1/2. But after you flip the coin, that probability doesn't change. The two events are independent. You can multiply them.

The only time you add them is if you are combining more than one probabilities into a single event. For example, if the probability I pick red is 1/4, and the probability I pick yellow is 2/4, then the probability I pick red OR yellow is 3/4.

Dependent events are a little harder to calculate. You do multiply them, but the probabilities change as you progress. You should watch this video to start with: https://www.khanacademy.org/math/precalculus/prob-comb/dependent-events-precalc/v/introduction-to-dependent-probability
• the probability of getting two heads in a row for a fair coin is 0.5*0.5. My question is why do we multiply both the probabilities? why dont we add 0.5 and 0.5?
(1 vote)
• The coin flips are independent events. That is, the first flip does not influence what the second flip will be in any way. So we see that there are 4 possibilities:
HT
TH
HH
TT
These possibilities could be counted by noting that we have 2 options for the first flip, but each option of the first flip can be associated with one of 2 different possible options for the second flip. Thus our total count of possibilities is 2 • 2 = 4 which we also found by listing them all out above. Remember the probability is desired/total, and we know that there is only 1 desired and 2 • 2 total, so we can express the probability as:
(1/2)(1/2) = 1/(2 • 2) = 1/4
I suggest you watch the videos on independent events and multiplying probabilities on KA to gain more intuition. Comment if you have questions!
• I also got a 0.2889 answer. I see below many others actually got the same answer as me but nobody actually answered why this was wrong or attended to correct it. Anyone has any idea?
• Great question!

The answer 0.2889, or more precisely 0.28890625 = (0.6*3/8 + 0.5*5/8)^2, is the result of misinterpreting the problem as selecting a coin, flipping it, putting it back, selecting a coin again, and flipping it. Equivalently, this is the result of mistakenly assuming that the two flips are overall independent.

The flips are not overall independent because the same coin is flipped twice, and if the first flip is heads, this increases the probability that the coin was the type that favors heads, which in turn increases the probability that the second flip of the same coin is heads. This is why the probability that both flips are heads is slightly larger than 0.28890625.

Have a blessed, wonderful day!