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    1. Dharmesh has 120 tins, of which 50 contained pet food, 20 contained peas, 35 contained beans and the rest contained soup. Find the probability that Dharmesh will have to open more than two tins before she finds one which does not contain pet food.

    The back of the book gives the answer as 35/204 but I have no idea how to get this answer. Is it something to do with binomial distribution or what?

    2. Three friends all have birthdays during January. Assuming that each friend’s birthday is equally likely to be on any of the 31 days of January, find the probability that all three friends’ birthdays are on different days.

    The back of the book gives an answer of 870/961 but I don’t know how to get this answer.

    Can you please help me with these questions because I am so horribly stuck
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    the second one:

    prob of 1st person having bday any day in jan = 1
    prob of 2nd person having a bday any day except one = 30/31
    prob of 3rd person having a bday any day except above 2 = 29/31

    1 x 30/31 x 29/31 = 870/961
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    (Original post by alison_141288)
    1. Dharmesh has 120 tins, of which 50 contained pet food, 20 contained peas, 35 contained beans and the rest contained soup. Find the probability that Dharmesh will have to open more than two tins before she finds one which does not contain pet food.

    The back of the book gives the answer as 35/204 but I have no idea how to get this answer. Is it something to do with binomial distribution or what?
    If Dharmesh has to open two tins before she finds one which does not contain pet food, she will open pet food tins on her first 2 choices.

    so P(2 pet foods in a row) = 50/120 x 49/119 = 35/204
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    cheers guys - it seems so simple when you explain. i have one more quick question:
    1) Describe how random number could be used to select a random sample of 10 of the greengrocer’s oranges if they have 240 oranges in stock.

    The back of the book says: number the oranges 000 to 239. Use a calculator to select numbers in the range. 3-digit numbers are chosen and numbers >239and repeats are ignored. Repeat until 10 are chosen

    However, I wrote that you could use the random number function on the calculator to generate a random number. Multiply that number by 239 to get a random number between 0 and 239. Ignore repeats and continue until 10 are chosen

    Is my answer still correct?
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    (Original post by alison_141288)
    cheers guys - it seems so simple when you explain. i have one more quick question:
    1) Describe how random number could be used to select a random sample of 10 of the greengrocer’s oranges if they have 240 oranges in stock.

    The back of the book says: number the oranges 000 to 239. Use a calculator to select numbers in the range. 3-digit numbers are chosen and numbers >239and repeats are ignored. Repeat until 10 are chosen

    However, I wrote that you could use the random number function on the calculator to generate a random number. Multiply that number by 239 to get a random number between 0 and 239. Ignore repeats and continue until 10 are chosen

    Is my answer still correct?
    yh sounds fine to me. You could type in shift ran#(0239) and that gives you the numbers without having to multiply by 239. I'd also change your calculator settings to 0 d.p to avoid rounding errors, and number oranges 0 to 239 as the book states.
 
 
 
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