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Math Help - Probability Question

  1. #1
    Member OnMyWayToBeAMathProffesor's Avatar
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    Probability Question

    Hello,

    I am stuck on this probability question:

    A fair coin is tossed 6 times. What is the probability of tossing exactly four heads in 6 tosses? What is the probability of tossing at most two heads. Each the first and second questions are independant of each other.

    I believe I would know how to do it if if asked for a dice, because then the probabailty for each number is different. But the probability of heads or ails is both 50%. This is where I get stuck.

    I would appreciate any help. Thank you.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by OnMyWayToBeAMathProffesor View Post
    Hello,

    I am stuck on this probability question:

    A fair coin is tossed 6 times. What is the probability of tossing exactly four heads in 6 tosses? What is the probability of tossing at most two heads. Each the first and second questions are independant of each other.

    I believe I would know how to do it if if asked for a dice, because then the probabailty for each number is different. But the probability of heads or ails is both 50%. This is where I get stuck.

    I would appreciate any help. Thank you.
    this is a binomial distribution problem.

    Suppose we have n independent trials, the outcome of which can be considered a success or failure, and we are looking for k successes each with a probability of p. we call the probability of failure q, where q = 1 - p, of course.

    the probability of k successes is given by: P(k) = {n \choose k}p^kq^{n - k}

    by the way, usually problems with dice are a bit more complicated than problems with coins
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  3. #3
    Member OnMyWayToBeAMathProffesor's Avatar
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    Reply to Jhevon

    So I am grateful for your formula and I put it to use. This is what i got:

    for the probability of tossing exactly 4 heads;

    P(4) = {6 \choose 4}(\frac{1}{2})^4(\frac{1}{2})^{6 - 4} and I received 23.4%. Is this correct?

    for the probability of at most two heads;

    ({6 \choose 1}(\frac{1}{2})^1(\frac{1}{2})^{6 - 1})+({6 \choose 2}(\frac{1}{2})^2(\frac{1}{2})^{6 - 2}) and I received 32.8%. Is this correct?

    Thank you.
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by OnMyWayToBeAMathProffesor View Post
    So I am grateful for your formula and I put it to use. This is what i got:

    for the probability of tossing exactly 4 heads;

    P(4) = {6 \choose 4}(\frac{1}{2})^4(\frac{1}{2})^{6 - 4} and I received 23.4%. Is this correct?
    yes


    for the probability of at most two heads;

    ({6 \choose 1}(\frac{1}{2})^1(\frac{1}{2})^{6 - 1})+({6 \choose 2}(\frac{1}{2})^2(\frac{1}{2})^{6 - 2}) and I received 32.8%. Is this correct?

    Thank you.
    no. of course you realize "at most two heads" means two or less. what you forgot is that zero heads is a possible outcome that is less than two heads
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