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27 January, 07:27

You have a jar of jelly beans in front of you

with 12 - lime, 17 - papaya, 5 - mango and 13

- bubble gum. What is the probability, as a

fraction, of selecting either a lime or bubble

gum followed by a papaya?

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Answers (1)
  1. 27 January, 09:27
    0
    Answer: The probability of selecting either a lime or bubble gum followed by a papaya is 425/2209

    Step-by-step explanation:

    Since, there are four types of items in the jar, namely - lime, papaya, mango and bubble gum that have quantities of 12, 17, 5 and 13 respectively. Before finding the probabilities, we first add up the quantities or number of different items in the jar.

    12 + 17 + 5 + 13 = 47 in total.

    Since lime is 12 in number, the probability of selecting a lime is = 12/47.

    Bubble gums are 13 in number, so the probability of selecting a bubble gum = 13/47.

    Then the probability of selecting either a lime or a bubble gum is the addition of the probabilities of selecting a lime and selecting a bubble gum:

    = 12/47 + 13/47

    = 25/47.

    Again, the probability of selecting a papaya is =

    number of papaya/total number of items

    = 17/47

    If the system is with replacement, then the probability of selecting a lime or bubble gum followed by a papaya is:

    = Probability of selecting a lime or bubble gum * Probability of selecting papaya

    = 25/47 * 17/47

    = 425/2209

    The probability of selecting a lime or bubble gum followed by a papaya is 425/2209
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