Someone has stolen an ATM card and knows that the first and last digits of the PIN are 8 and 1, respectively. He has three tries before the card is retained by the ATM (but does not realize that). So he randomly selects the 2nd and 3rd digits for the first try, then randomly selects a different pair of digits for the second try, and yet another randomly selected pair of digits for the third try (the individual knows about the restrictions described in (b) so selects only from the legitimate possibilities). What is the probability that the individual gains access to the account

The probability of the individual to gain access into the account is 0.038

Step-by-step explanation:

The security code of every ATM card is four digits,

So let's assume the code is

8xy1

Let x and y represent the second and third digit

STEP 1 : THE PROBABILITY OF GETTING ACCESS IN THE FIRST TRIAL

to select a number from 0 to 9 is ten chances

Therefore the probability to select the right number will be;

1/10

The probability of the right value of x and y to occur the same time will be

1/10 × 1/10 = 1/100

STEP 2: THE PROBABILITY OF GETTING ACCESS IN THE SECOND TRAIL;

Since a number has been selected in the first trail, the probability of selecting the right number will be; 1/9

The probability of getting the right value of x and y to occur the same time will be

1/9 × 1/9 = 1/81

STEP 3: THE PROBABILITY OF GETTING ACCESS IN THE THIRD TRIAL;

Since a number has been selected in the first and second trail, the probability of selecting the right number will be;

1/8

The probability of getting the right value of x and y to occur the same time will be;

1/8 × 1/8 = 1/64

STEP 4: THE PROBABILITY OF GETTING ACCESS TO THE ACOUNT;

For the person to gain access to the account that means the first, second, and their probability is applied.

1/100 + 1/81 + 1/64 = 19684/518400 = 0.038

Therefore the probability of getting access into the account is 0.038

final answer is : )