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I have just extracted from 2 consecutive tries the same ticket out of 4.

How do I calculate the probability of such an event?

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  • $\begingroup$ are you extracting the tickets from an urn? are the tickets equally likely to be selected? $\endgroup$ – Conrado Costa Sep 3 '15 at 13:25
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If you're looking for any ticket twice: Whichever ticket you pick the first time, you have a 1/n (in this case 1/4) chance of it being the ticket that you pick next time

If you're looking for a specific ticket twice: 1/4 the first time * 1/4 the second time = 1/16

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First time you picked one ticked out of 4. i.e 1/4 =no of favorable/total no of outcomes. Again you picked the same ticket out of four i.e 1*4 *1*4 =1/16

i.e P=1/16

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The probability of drawing a specific ticket out of 4 is 1/4. You are drawing with replacement. Hence, the probability that the next ticket drawn is the same ticket is also 1/4. To get the probability of the event, you multiply the probability of getting the specific ticket in the first draw with the probability of getting the same ticket on the second draw. i.e. 1/4 * 1/4 in this case

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You are looking at joint probability. $P(A \cdot A) = P(A)\cdot P(A|A)$. Where $P(A)$ is the probability to draw the ticket of interest. Since you are drawing with replacement you have the conditional probability $P(A|A)$ reducing to simple probability $P(A)$. On each draw you have 4 ways of choosing the ticket and hence $P(A)=\frac{1}{4}$. Which gives for $P(A \cdot A)=\frac{1}{4}\cdot \frac{1}{4}$.

Remark: Notice that if you had drawing without replacement you would not have the case $P(A|A)=P(A)$!

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