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anonymous

  • one year ago

A football quarterback has 2 more chances to throw a touchdown before his team is forced to punt the ball. He misses the receiver on the first throw 30% of the time. When his first throw is incomplete, he misses the receiver on the second throw 10% of the time. What is the probability of not throwing the ball to a receiver on either throw?

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  1. anonymous
    • one year ago
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    3/20 would be the right answer I think, but i don't know for sure.

  2. anonymous
    • one year ago
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    or 15%

  3. kropot72
    • one year ago
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    Let A be the event he misses on the first throw. P(A) = 0.3 Let B be the event he misses on the second throw. Then P(B|A) = 0.1 \[\large P(A\cap B)=P(A)\times P(B|A)=0.3\times0.1=0.03\]

  4. kropot72
    • one year ago
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    If the probability of not throwing the ball to a receiver on either throw is expressed as a percentage it is 0.03 * 100 = 3%.

  5. anonymous
    • one year ago
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    are you sure that is right how did you get that? @Pencera

  6. anonymous
    • one year ago
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    no im not i said that in my answer

  7. anonymous
    • one year ago
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    i apologize

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