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stiks02 [169]
2 years ago
6

Contestants on a game show spin a wheel with 242424 equally-sized segments. Most of those segments show different prize amounts,

but 222 of them are labeled "bankrupt":Suppose that a contestant is going to spin the wheel twice in a row, and the outcome of one spin doesn't affect the outcome of future spins. What is the probability that NEITHER of the spins land on "bankrupt"?
Mathematics
1 answer:
ycow [4]2 years ago
3 0

Using the principle of probability, the probability that the outcome of both spins does not land on "<em>bankrupt</em><em>"</em><em> </em>is 1/144

<u>Given</u><u> </u><u>the</u><u> </u><u>Parameters</u><u> </u><u>:</u>

  • Total Number of possible outcomes = 24
  • Number of outcomes labeled bankrupt = 2
  • Labels which aren't labeled bankrupt = 24 - 2 = 22

<u>Recall</u><u> </u><u>:</u>

  • P(A) = <em>required</em><em> </em><em>outcome</em><em> </em><em>/</em><em> </em><em>Total</em><em> </em><em>possible outcomes</em><em> </em>

<u>First</u><u> </u><u>spin</u><u> </u><u>:</u>

  • P(not bankrupt) = 2 / 24 = 1/12

<u>Second</u><u> </u><u>spin</u><u> </u><u>:</u>

  • P(not bankrupt) = 2/24 = 1/12

P(<em>neither</em><em> </em><em>lands</em><em> </em><em>on</em><em> </em><em>bankrupt</em><em>)</em><em> </em><em>=</em><em> </em>1/12 × 1/12 = 1/144

Therefore, the probability that neither lands on bankrupt is 1/144

Learn more : brainly.com/question/18405415

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