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If a certain coin is flipped

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rsarashi Master | Next Rank: 500 Posts Default Avatar
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If a certain coin is flipped

Post Sat Sep 02, 2017 11:42 am
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    If a certain coin is flipped, the probability that the coin will land heads up is 1/2. If the coin is flipped 5 times, what is the probability that it will land heads up on the first 3 flips and not on the last 2 flips?

    A) 3/5

    B) 1/2

    C) 1/5

    D) 1/8

    E) 1/32

    OAE

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    Post Sat Sep 02, 2017 11:49 am
    rsarashi wrote:
    If a certain coin is flipped, the probability that the coin will land heads up is 1/2. If the coin is flipped 5 times, what is the probability that it will land heads up on the first 3 flips and not on the last 2 flips?
    A) 3/5
    B) 1/2
    C) 1/5
    D) 1/8
    E) 1/32

    OAE
    On ANY individual toss, P(heads) = 1/2 and P(tails) = 1/2

    P(heads 1st toss AND heads 2nd toss AND heads 3rd toss AND tails 4th toss AND tails 5th toss AND)
    = P(heads 1st toss) x P(heads 2nd toss) x P(heads 3rd toss) x P(tails 4th toss) x P(tails 5th toss)
    = 1/2 x 1/2 x 1/2 x 1/2 x 1/2
    = 1/32

    Answer: E

    Cheers,
    Brent

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    Post Sun Sep 03, 2017 3:45 am
    rsarashi wrote:
    If a certain coin is flipped, the probability that the coin will land heads up is 1/2. If the coin is flipped 5 times, what is the probability that it will land heads up on the first 3 flips and not on the last 2 flips?

    A) 3/5

    B) 1/2

    C) 1/5

    D) 1/8

    E) 1/32

    OAE
    I'm surprised that there is no answer as 5/16. This makes someone safe in case he/she does not get the question well.

    Brent explained this one quite well.

    Had the question been...

    Quote:
    If a certain coin is flipped, the probability that the coin will land heads up is 1/2. If the coin is flipped 5 times, what is the probability that it will land heads up on 3 flips and tail on 2 flips?
    Unlike the original question in which the condition to get heads on the first 3 flips and tail to get on the last 2 flips, in the above-modified question, there is no such condition, thus, only important is to get 3 head out of 5 flips, be it in any order. A couple of instances HTHTH and HHTTH are valid.

    Three instances of heads out of five instances is given by 5C3 = (5.4.3) / (1.2.3) = 10 ways.

    Thus, the required probability = 5C3*[(1/2*1/2*1/2)*(1/2*1/2)] = 10*[1/2^5] = 5/16.

    Hope this helps!

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