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sh3lsh

  • one year ago

A coin is biased so that the probability of heads is 2/3. What is the probability that exactly four heads come up when the coin is flipped seven times, assuming that the flips are independent?

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  1. sh3lsh
    • one year ago
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    C(7, 4)(2/3)^4(1/3)^3 is the answer. Could someone explain how they reached this conclusion?

  2. kropot72
    • one year ago
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    The binomial distribution applies in this case. Have you studied the binomial distribution?

  3. sh3lsh
    • one year ago
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    It does! Thanks, I got it now!

  4. sh3lsh
    • one year ago
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    On a side note, could you help me with this?

  5. sh3lsh
    • one year ago
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    Given a coin that has a head 3 times more likely to come up than a tail, what's the probability of getting a head and a tail?

  6. sh3lsh
    • one year ago
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    I think the solution is (3/4) (1/4) + (1/4) (3/4) = 6/16 , but I'm not sure.

  7. kropot72
    • one year ago
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    This question is not clear. However if it is asking for the probability of getting a head first and then a tail on two successive tosses, then the calculation would be as follows: On the first toss, P(head) is 3/4. On the second toss, P(tail) is 1/4. The events 'head on first toss' and 'tail on second toss' are independent, meaning the outcomes have no effect on each other. Therefore P(head on first toss, tail on second toss) = 3/4 * 1/4 = 3/16

  8. sh3lsh
    • one year ago
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    I agree, and I think that's true.

  9. kropot72
    • one year ago
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    However if the order in which a head and a tail were obtained does not matter, the result would be as you wrote it.

  10. sh3lsh
    • one year ago
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    If one was to ask the probability of getting two head when tossing two of these coins, would you add 3/4 + 3/4 ?

  11. kropot72
    • one year ago
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    No. The results of each toss are independent, therefore the probabilities of heads must be multiplied. This gives 3/4 * 3/4 = 9/16.

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