Hi,
So I need some help:
A customer wishes to withdraw money from a cash machine. To do this it is necessary to type a PIN number into the machine. The customer is unsure of this number. If the wrong number is typed in, the customer can try again up to a maximum of four attempts in total. Attempts to type in the correct number are independent and the probability of success at each attempt is 0.6.
The random variable A represents the number of attempts made to type in the correct PIN number, regardless of whether or not the attempt is successful.
(b) Find the probability distribution of A.
Ok, so my probability distribution:
a: 1, 2, 3, 4
P(A=a): 0.6, 0.24, 0.096, 0.0384
But the mark scheme is different for P(A=4), which it says it is 0.064.
I don't understand why it would be 0.064 as I thought it would be:
fail*fail*fail*success
Thanks.