Question

ATM You want to obtain cash by using an ATM, but it’s dark and you can’t see your card when you insert it. The card must be inserted with the front side up and the printing configured so that the beginning of your name enters first.

a. What is the probability of selecting a random position and inserting the card with the result that the card is inserted correctly?

b. What is the probability of randomly selecting the card’s position and finding that it is incorrectly inserted on the first attempt, but it is correctly inserted on the second attempt? (Assume that the same position used for the first attempt could also be used for the second attempt.)

c. How many random selections are required to be absolutely sure that the card works because it is inserted correctly?

Short answer

a.The probability of correctly inserting the card is equal to 0.25.

b.The probability of inserting the card incorrectly at the first attempt and correctly at the second attempt is equal to 0.1875.

c.If the card is placed wrongly in the first three attempts, the card will almost certainly be inserted correctly at the fourth attempt because there are only four ways by which a card can be inserted. As a result, the number of choices required to be completely certain that the card is inserted correctly is equal to 4.

Step-by-step solution

Given information

In order to collect cash from the ATM, the card must be inserted with the front side up and the printed name should enter first.

Probabilities

a.

The four possible ways of inserting the card inside the machine are written below:

• Front side up and the beginning of the name is inserted first.
• Front side up and the beginning of the name is not inserted first.
• Back side up and the beginning of the name is inserted first.
• Back side up and the beginning of the name is not inserted first.

Thus, the total number of ways of inserting the cardis equal to 4.

The number of ways in which the correct side of the card is inserted with the printed name being inserted first is equal to 1.

The probability of correctly inserting the card is computed below:

$$\begin{gathered} P\left(\text{correctly}\text{inserting}\text{the}\text{card}\right)=\frac{1}{4} \\ =0.25 \end{gathered}$$

Therefore, the probability of correctly inserting the card is equal to 0.25.

b.

The probability of inserting the card incorrectly is given below:

$$\begin{gathered} P\left(\text{incorrectly}\text{inserting}\text{the}\text{card}\right)=1-P\left(\text{correctly}\text{inserting}\text{the}\text{card}\right) \\ =1-\frac{1}{4} \\ =\frac{3}{4} \\ =0.75 \end{gathered}$$

Now, the probability of inserting the card incorrectly at the first attempt and correctly at the second attempt is given below:

$\left(0.75\right)\left(0.25\right)=0.1875$

Therefore, the probability of inserting the card incorrectly at the first attempt and correctly at the second attempt is equal to 0.1875.

c.

There are a total of 4 ways to insert the card, out of which only 1 way is correct.

If the card is inserted incorrectly in the first 3 attempts, then it is sure that the card will be inserted in the correct wayat the 4^thattempt.

Thus, it can be inferred that the number of selections required to be absolutely sure that the card is working because it was inserted correctly is equal to 4.