Discrete probability distribution question (candy wrapping)

[Monkeyseat]

Hi,

Question

The probability distribution for the number, R, of unwrapped sweets in a tin is given in the following table:

a) Show that:
i) E(R) = 2,
ii) Var(R) = 1.2.

b) The number, P, of partially wrapped sweets in a tin is given by P = 3R + 4. Find values for E(P) and Var(P).

c) The total number of wrapped sweets in a tin is 200. Sweets are either correctly wrapped, partially wrapped or unwrapped.
i) Express, C, the number of correctly wrapped sweets in a tin in terms of R.
ii) Hence find the mean and variance of C.

Working

ai) No problem with this.
aii) No problem with this.

b) E(P) = 10 and Var(P) = 10.8

ci) This is the part I'm not sure about.

There are 200 wrapped sweets in the tin. "3R + 4" is the amount that are partially wrapped.

Therefore the number of correctly wrapped sweets is given by:

C = 200 - (3R + 4)
C = 196 - 3R

The book says the answer for this part the question is C = 196 - 4R. I don't know where this came from. By the looks of it, they did C = 200 - (3R + 4 + R), but R is the number of unwrapped sweets, so I thought it is irrelevant...

The number of correctly wrapped sweets = The number of wrapped sweets - the number of partially wrapped sweets
C = 200 - (3R + 4)

How does the number of unwrapped sweets come into this? How did the book get the answer that it did?

Thanks.

 

[royhaas]

Re: Discrete probability distribution question

You need to add the partially wrapped to "totally" unwrapped to get the total "defective" count.

 

[Monkeyseat]

Re: Discrete probability distribution question

But 200 is the amount that is either correctly wrapped or partially wrapped (this does not include the amount that is unwrapped). So if I take "3R + 4", this being the amount that is partially wrapped, away from 200, this should give me the amount that is correctly wrapped. So, C = 196 - 3R.

That makes sense to me, but it is wrong. Why? Does 200 include the amount that is unwrapped?

Thanks.

 

[Deleted member 4993]

Re: Discrete probability distribution question

Question

The probability distribution for the number, R, of unwrapped sweets in a tin is given in the following table:

a) Show that:
i) E(R) = 2,
ii) Var(R) = 1.2.

b) The number, P, of partially wrapped sweets in a tin is given by P = 3R + 4. Find values for E(P) and Var(P).

c) The total number of wrapped sweets in a tin is 200. Sweets are either correctly wrapped, partially wrapped or unwrapped.

200 = C + P + R

 

[Monkeyseat]

Subhotosh Khan, I know that sweets can be correctly wrapped, partially wrapped or unwrapped, but I still don't see why 200 = C + P + R. Bear with me please.

200 is the number of wrapped sweets in a tin. If these 200 sweets are wrapped, they must all be either correctly wrapped or partially wrapped i.e. 200 = C + P. So C = 200 - P.

What I am trying to say is, if 200 is the number of sweets in a tin and they are all wrapped, they must all be either correctly wrapped or partially wrapped - there can be no unwrapped sweets if all 200 sweets are wrapped.

Why does the 200 wrapped sweets include unwrapped sweets?

Thanks.