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Probability calculation is about determining how likely an event is to occur. In this exercise, we are calculating the probability of inserting an ATM card correctly. When we talk about probability, we use fractions to express it. For example, there are four different ways to insert the card, but only one correct way. So, the probability of inserting it correctly the first time is \(\frac{1}{4}\). To find this, we divide the number of correct outcomes by the total number of possible outcomes.

For calculating the probability of incorrect insertion the first time and correct insertion the second time, we multiply the probabilities of each outcome. If the card is inserted incorrectly on the first attempt, the probability is \(\frac{3}{4}\). If the card is then inserted correctly on the second attempt, the probability is \(\frac{1}{4}\). We multiply these together to get \(\frac{3}{16}\).

Probability helps us understand the chances of different outcomes and make informed decisions.

Random selection means choosing something without any particular order or pattern. In the case of the ATM card insertion problem, each attempt to insert the card is a random selection. We do not know the correct orientation, so each of the four possible positions is equally likely.

This randomness is why we use probability to determine the likelihood of the card being inserted correctly. Whether we choose the front side up or back side up, and whether the name is first or last, are all treated as random choices. Using randomness, we can manage expectations and understand the possible outcomes.

The ATM card insertion problem illustrates how probability and random selection work in real-life situations. When it's dark, and we can't see how to insert the ATM card, we have four possible ways to insert it: front side up with the name entering first, front side up with the name entering last, back side up with the name entering first, and back side up with the name entering last.

a. The probability of inserting the card correctly on the first try is \(\frac{1}{4}\), because only one out of the four possible positions is correct.
b. If the card is inserted incorrectly on the first try, the probability is \(\frac{3}{4}\). For it to be correct on the second try, we again have a probability of \(\frac{1}{4}\). Multiplying these probabilities together gives \(\frac{3}{16}\).
c. To be absolutely sure the card is inserted correctly, we must try all four possible positions. As long as we try all four, one of them has to be correct.

This simple problem helps us understand the broader concepts of probability and randomness in real-world scenarios. Knowing these concepts can help us make better predictions and decisions in uncertain situations.