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Independence and dependence with events like first card red and second card black and of course, without replacement. I believe they are dependent but have not found the best way to explain this. A few of my very bright students in discrete math believe these events are independent but I think they are dependent. Can anyone help me to give a good, concise explanation?

meldolly
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  • If we take the first red card away, does it increase or decrease the chance getting a black card ? If we had 3 red cards, 4 black cards, and 5 green cards, for example, does the probability of drawing a black card change when we draw a red card ? – Vivek Kaushik Mar 28 '17 at 01:43
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    As an aside, for a standard playing card deck, the probability that the first card is red is equal to the probability that the second card is black and is $\frac{1}{2}$. Had the events been independent, then the probability of the intersection would have been $\frac{1}{4}$ however the probability is in fact $\frac{26\cdot 26}{52\cdot 51}$ – JMoravitz Mar 28 '17 at 02:19

1 Answers1

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You are correct that the two events are dependent.

If the first cards is red, then $26$ of the remaining $51$ cards are black, so the probability of the second card being black is $26\over 51$.

If the first cards is not red, then $25$ of the remaining $51$ cards are black, so the probability of the second card being black is $25\over 51$.

Since the probability of the second card being black changes based on whether or not the first card is red, the two events are dependent.

browngreen
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  • This is my reasoning too, but my textbook suggests a different reason. https://math.stackexchange.com/q/4403670/306540 – B flat Mar 15 '22 at 19:31