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S1 question: What the hell is it talking about?!?!?!? watch

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    Hi, first I would like to say that 'EDEXCEL are $h!t'.

    Now, I have an S1 resit tomorrow morning. I got 58 last year so I will definitely need to resit. I'm currently going through the Heinemann's Revise for Statstics 1 book and there's a question on (for those who have it) page 60. It's questions (a) and (b), with (a) as a proof question and then (b) as working out the general probability distribution for X. Anyways, here it goes:

    When a certain type of cell is subjected to radiation, the cell may die, survive as a single cell, or divide into two cells with probabilities 1/2, 1/3, 1/6 respectively.

    Two cells are independently subjected to radiation. The random variable X represents the total number of cells in existence after this experiment.

    (a) Show that P(X=2) = 5/18
    (b) Find the probability distribution of X

    I have tried all sorts of things wth (a). I did manage to get 5/18 once but I'm pretty sure it wasn't how you work it out. Stupid textbook never gives answers for proof questions. I implemented my method to (b) and it's not right.

    Here are the answers for (b):

    x: 0 1 2 3 4
    p(x): 9/36 12/36 10/36 4/46 1/36

    So can anyone help me then because I absolutely have NO IDEA WHAT IS GOING ON. This is why I hate Statistics because some questions blab on and on and I have no-idea what it's talking about. Why can't the questions be straightforward? The exam boards should realise that not everyone is quick at realising stuff and it's about getting the correct answer, not about getting it as quick as possible. I would probably get the answer after looking at it for a few hours but I still need to do a past paper or two before I finish tonight.
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    For a): The probability that there will be 2 cells left at the end is the sum of the probabilities that they both survive and that one dies and the other divides. So:

    Probability that they both survive = (1/3)^2 = 1/9

    Probability that one dies and the other divides = (1/2)*(1/6) + (1/2)*(1/6)
    = 1/6

    1/9 + 1/6 = 5/18

    Ben
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    (Original post by Ben.S.)
    For a): The probability that there will be 2 cells left at the end is the sum of the probabilities that they both survive and that one dies and the other divides. So:

    Probability that they both survive = (1/3)^2 = 1/9

    Probability that one dies and the other divides = (1/2)*(1/6) + (1/2)*(1/6)
    = 1/6

    1/9 + 1/6 = 5/18

    Ben
    Oh I see! I thought that the probability that they either remain as one cell or divde already means that they will stay alive! They should word these questions better! Thanks man!
 
 
 
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