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# Probability of Dependent Events - Problem 1

###### Carl Horowitz

###### Carl Horowitz

**University of Michigan**

Runs his own tutoring company

Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!

Finding the probability of dependent events or events that one affects the outcome of another. So what we're at here is the probability of being dealt blackjack.

So for this one we're going to make some assumptions that it's just you versus the dealer and the dealer keeps both your cards faced down, so really the only card that we know is going to be out there is that first card, everything is going to remain unknown.

So for those of you who don't know black jack is basically card game where one of the objects is to try to get an ace and a face card or a 10, so basically they add up to 21 because ace can count as 11. So in order to have blackjack what we need is an ace, so we say we are dealt an ace so there are 4 aces in a deck, there is 52 cards in a deck so we have a 4 out of 52 chance of being dealt an ace.

So once our first card is an ace, we now need a 10 jack queen or king to finish up our hand, so there are 4 cards that are satisfied, but there's also 4 suits, so there are 16 cards that we were on and there are only 51 cards left in the deck that we know remember the aces face up so our cards are out every other card is unknown, so we're just left with 51 cards, so we have times 16 over 51.

This one actually is a little bit of a trick, so this one we assumed we're getting the ace first and then the face card, but order doesn't necessarily matter, we could be dealt the face card first, so what we have to do is actually add assuming we get the face card first which is going to be 16 out of 52 and then the ace which is 4 out of 51. Basically the same thing as this, we just flipped our numerators, so really what we end up with is 4 out of 52 times 16 out of 51 times 2.

So finding the probability of dependent events sort of think about the probability of 1 and then think about how that outcome will affect the other. In this case we had 52 cards at the start, for our second event we only had 51 left so therefore it's dependent because the first event changed the outcome for the second.

Pretty straight forward stuff, but just think about it a little more and how one affects the other.

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###### Carl Horowitz

B.S. in Mathematics University of Michigan

He knows how to make difficult math concepts easy for everyone to understand. He speaks at a steady pace and his step-by-step explanations are easy to follow.

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