What is the probability of obtaining four tails in a row when flipping a coin interpret this probability?

There are several possible approaches.

Getting 4 heads to start has probability $(1/2)^4 = 1/16$ as in the comment by @DonatPants.

More formally, outcomes that satisfy your condition are HHHHT, HHHHHT, HHHHHHT, etc. So the total probability is the geometric series with probability $A = (1/2)^5 + (1/2)^6 + (1/2)^7 + \dots .$

There is a formula for summing this series. If you don't know it, you can note that $(1/2)A = (1/2)^6 + 1/2)^7 + \dots,$ so that $A - (1/2)A = = (1/2)A = (1/2)^5$ and $A = (1/2)^4,$ which is the same as the previous answer.

I do not know why @MarkusStuhr has withdrawn his Answer. The explanation by @Joel (+1) as does the answer by @manmood (+1) that appeared while was typing this. I hope one of these explanations is clear to you. The key points throughout is that we're assuming the coin is fair [$P(H) = 1/2$] and that the tosses are independent.

Also, if your book includes the geometric distribution, you should look at that because it is related to this problem. I don't want to discuss the geometric distribution in this Answer because there are at least two versions of it, and discussing the wrong one might be confusing.

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