Edexcel S1 Mock Paper question help

Hey, I'm really stuck on a question on the S1 mock paper. I can't find a copy of the marks scheme anywhere on the interent. Could osmone point me in the right directions?

Althernatively, could somone help me with it?

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. (2 marks)

(c) Calculate the probability that the customer types in the correct number in four or fewer attempts. (2 marks)

It's really annoying, because the next question I can do, but it relies on your getting this question right.

If anyone could get back to me asap it would be much appreciated. My exam is Wednesday :s-smilie:

Reply 1

screenager2004

14

I think we are going to have to help each other! I can answer the question, but what's the "Probability distribution" - In the books I learn from, I have the "Probability Density function" and a "Cumulative distribution function" but nothing that's just called a "probability distribution" - that's an entire chapter heading, its so vague so I don't really understand what they're asking for?!

Anyhoo, as for the actual numbers: We are looking at the rule P(a∩b) = P(a) x P(b), where a is a success and b is a failure.

P(a∩b) means The probability of two independent events "a" AND "b" occurring in the same experiment.

I would draw one of those sexy little trees but I have no idea how to make one with LaTeX or ASCII.

The chance of getting it right first time is 0.6
the chance of getting it right second time is 0.4 x 0.6 = 0.24
The chance of getting it right on the second time is 0.4 x 0.4 x 0.6 = 0.096
The chance of getting it right on the third attempt is 0.4 x 0.4 x 0.4 x 0.6 = 0.0384
The chance of getting it wrong every time is 0.4 x 0.4 x 0.4 x 0.4 = 0.0256

These are the only possible outcomes, and 0.6 + 0.24 + 0.096 + 0.0384 + 0.0256 = 1!

So the probability that you get it right in four attempts or less is 1-0.0256 = 0.9744

You'll have to forgive my ignorance. I never had formal maths lessons!

Reply 2

____Panic

Hi, I'm sorry I can't help with this because I'm also really confused. May I ask what does 'regardless whether or not the attempt is successful' mean, does it mean at the end the four attempts could be all wrong?

Original post by jazzyyazzy6

Hey, I'm really stuck on a question on the S1 mock paper. I can't find a copy of the marks scheme anywhere on the interent. Could osmone point me in the right directions?

Althernatively, could somone help me with it?

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. (2 marks)

(c) Calculate the probability that the customer types in the correct number in four or fewer attempts. (2 marks)

It's really annoying, because the next question I can do, but it relies on your getting this question right.

If anyone could get back to me asap it would be much appreciated. My exam is Wednesday :s-smilie:

Reply 3

Alpha-tega

2

Original post by ____Panic

Hi, I'm sorry I can't help with this because I'm also really confused. May I ask what does 'regardless whether or not the attempt is successful' mean, does it mean at the end the four attempts could be all wrong?

•

15-05-2009 01:36

8 years later... Legend has it that it was never solved.

Edexcel S1 Mock Paper question help

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