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20 February, 19:44

40% of OatyPop cereal boxes contain a prize. Hanna plans to keep buying cereal until she gets a prize. What is the probability that Hannah only has to buy 3 or less boxes before getting a prize? If we use a random number generator ranging from 1-10 and assign 1-4 as the prize and 5-10 as no prize, what is the BEST way to perform the simulation? a. keep track of how many times you see a 1,2,3 or 4 in 150 tries. These will represent the number of prizes you get in all. b. keep track of how many tries it takes before you see a 1,2,3 or 4. These represent the number of purchases it will take to get a prize. Repeat this 150 times. c. keep track of how many tries it takes before you see a 1,2,3 or 4. These represent the number of purchases it will take to get a prize. Repeat this 3 times.

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  1. 20 February, 19:48
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    keep track of how many tries it takes before you see a 1,2,3 or 4. These represent the number of purchases it will take to get a prize. Repeat this 150 times, option b
  2. 20 February, 20:10
    0
    She has a 88.4% chance to win.

    The correct method would be c.

    Step-by-step explanation:

    In order to determine whether she will win in 3 boxes or less, note that there is a 60% chance she will lose each time. To find the chance of her losing 3 times in a row, multiply 60% by itself 3 times.

    60% * 60% * 60% = 21.6%

    Now to find the amount of chance that she wins before this, subtract this number form 100%

    100% - 21.6% = 88.4%
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