# S1 Discrete random variablesWatch

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#1
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.
0
5 months ago
#2
(Original post by Wow_wizz)
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.
fail,fail,fail,fail is also included in A=4 since that also means that they attempted 4 times.
Last edited by Notnek; 5 months ago
0
5 months ago
#3
(Original post by Wow_wizz)
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.
note that the last one does not have to be correct !
Last edited by the bear; 5 months ago
0
#4
(Original post by Notnek)
fail,fail,fail,fail is also included in A=4 since that also means that they attempted 4 times.
Ok, thanks.
However, what is stopping for example P(A=2) from being 0.16 as it fail*fail which is still means it is attempted twice?

(Original post by the bear)
note that they last one does not have to be correct !
How would I know? Is it because they have to add to 1?
0
5 months ago
#5
(Original post by Wow_wizz)
Ok, thanks.
However, what is stopping for example P(A=2) from being 0.16 as it fail*fail which is still means it is attempted twice?
You have to assume that they keep trying until they are either successful or they fail 4 times. So fail, fail by itself could not happen because they would attempt at least one more time. Without this assumption, the question is not possible since we have no way of knowing when they will give up. The question should have been clearer.
Last edited by Notnek; 5 months ago
1
5 months ago
#6
" the customer can try again up to a maximum of four attempts"

it does not say any of the attempts have to be successful
0
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