Q. The digital security code needed to open a safe in a bank is 7835. A thief tries to open the safe by entering a randomly chosen sequence of 4 digits ( each digit is independent and equally likely to be any number from 0 to 9 ) A. Find the probability that the thief enters the correct code. B. Find P that the thief gets first or 2nd digit right (or both) C. Thief carries on entering digits till he gets it right, forgets which sequences he entered so all sequences are equally likely and successive attempts are independent. Let Y be the number of attempts till he enters the correct code. Name the distribution of Y and find P ( y>= 200) D. Suppose thief knows that the code is composed of digits 2,5,7,8, he enteres them in random orders. i) find P of thief entering right code. ii) Find p of entering exactly 2 of 4 digits in correct position. Attempt at Ansers : A: Probability of entering correct code = (1/10)^4 = 1/10,000 B: P of getting first digit right = 1/10, second = 1/10 so first or 2nd = 2/10 or 1/5, getting both right = 1/10*1/10 = 1/100 so, B = 20/100 + 1/100 = 21/100 C: I'm not sure what to call this distribution. Not uniform, binomial, exponential? Geometric I think but I can't explain why, I kind of don't understand exponential/geometric distributions. P(y>=200) erm I'm stumped. I know in a geometric is p x=k is q^k-1 * p where q is probability of first trial failing but I dont know how to go from x=k to x>k or x<k. Very confused and any input would be appreciated, thank you. D i) Correct code = 1/4! = 1/24 ii) Exactly 2 in correct order = ( 1/4 * 1/3 ) * 2, because you can arrange any 2 numbers in 2 ways. = 1/6.