OCR S2 question

Sorry for the loads of questions but I'm desperately trying to learn S2 before a mock this week

I can't seem to understand this question:
A hotel has 10 rooms, which are always occupied, and in each room there is a complimentary packet of biscuits. If on any one night the packet is opened, it is replaced by a new packet the next day. Probability of a packet being opened is 0.4. Manager wants to be at least 95% sure that he has enough packets in stock to replace the opened ones. Find the smallest number of packets he needs

Thank you

Don't fret it! This is what the forum is for. Also I might be getting a bit annoying, sorry...



So let's have $X \sim B(10, 0.4)$, you want to find $P(X \leq n) \geq 0.95$. Make sense?

A hotel has 10 rooms

If on any one night the packet is opened, it is replaced by a new packet the next day.

Probability of a packet being opened

is 0.4.

Manager wants to be at least 95% sure that he has enough packets in stock to replace the opened ones. Find the smallest number of packets he needs

I was trying to do >= n for some reason thank you so much, again!

Only 4 out of 10 will eat the biscuits? OCR, you crazy?

For some clarification:



Should be making you think binomial.



Keeps the number of trials constant.



Two outcomes, success/failure: packet being opened/packet not being opened/




Probability is constant.



Cumulative binomial.

This is great, thank you!

I have another question that I am a bit confused over, can I send it to one of you? It's from a past paper

I have another question that I am a bit confused over, can I send it to one of you? It's from a past paper

Can one of you help me with the last part of Q9 from Jan 13? I should know how to do this but I don't http://www.ocr.org.uk/Images/144610-question-paper-unit-4733-01-probability-and-statistics-2.pdf

@tinkerbella~ @Zacken @other clever maths people

Can one of you help me with the last part of Q9 from Jan 13? I should know how to do this but I don't http://www.ocr.org.uk/Images/144610-question-paper-unit-4733-01-probability-and-statistics-2.pdf

@tinkerbella~ @Zacken @other clever maths people

I think I should probably go into some more detail, actually:

Draw a tree diagram, the first branch is a split into $p = 0.5$ (with probability 0.2) or $p = 0.6$ (with probability 0.8).

The second split is Reject H0/Do not reject H0 (with probabilities that you should have found previously).

The third split is again Reject H0/Do not reject H0, but remember that $p$ is adjusted to 0.6 if the null hypothesis was rejected the first time around in the 0.5 branch.

Then just multiply along the tree and add the relevant probabilities.

That's really helpful, thank you so much! I need to start using tree diagrams more often

Does it work?

I think so, I got 0.083 or something but I haven't checked the mark scheme yet

Let me know when you do!