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anonymous

  • 5 years ago

A fair die is tossed 4 times. Find the probability of a. at least 2 5s b. at most 3 even numbers c. no more than 1 odd number d. at least 1 prime number e. at most 1 number greater than 4

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  1. Yuki
    • 5 years ago
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    to find a), you need to know that P(at least 2 5's) = P(exactly 2 5s) + P(exactly 3 5s) +P(exactly 4 5s)+P(all 5s)

  2. Yuki
    • 5 years ago
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    so you can find each of those probabilities and add them. that would be the straight forward way.

  3. Yuki
    • 5 years ago
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    the easier way is to consider the complement of P(at least 5) which is P(at most 1)

  4. Yuki
    • 5 years ago
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    that way all you have to do is calculate P(at most 1 5s) = P(no 5) + P(exactly 1 5)

  5. Yuki
    • 5 years ago
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    Then you can use the fact that P(event A) = 1 - P( complement of event A)

  6. Yuki
    • 5 years ago
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    does that help you at all ?

  7. Yuki
    • 5 years ago
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    From the looks of it, all problems I see here becomes much easier if you use P(A) = 1-P(notA) let me know if you still need help

  8. anonymous
    • 5 years ago
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    how about the other parts?

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