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Math - Calculating probability

JohnDoeJohnDoe Registered User regular
edited March 2010 in Help / Advice Forum
I'm trying to calculate the probability of this situation
I have a list of 10 numbers (1 through 10). The list is then randomised.
What are the odds of the first 4 numbers containing
- 1 and either 2 or 3 or 4

I've calculated this as (4 Choose 2) * 1/10 (for the 1) * 3/10 (for 2 through 4). Which gives me 18%. But I wrote a little random number number generator and the results don't match up (over a large distrubtion). Where am I going wrong with this?

JohnDoe on

Posts

  • .kbf?.kbf? Registered User regular
    edited March 2010
    If it's and you multiply. If it's or you add.

    So you want 1 and (2 or 3 or 4). The probability any one of these showing up is 1/10

    so (1/10)[(1/10)+(1/10)+(1/10)] = (1/10)(3/10)=3/100

    I think. I took prop and stats a long time ago...

    .kbf? on
  • JohnDoeJohnDoe Registered User regular
    edited March 2010
    The thing is, theres 4 spots where the 1 and 2/3/4 can appear. So how do they get worked into the calculation?

    JohnDoe on
  • PlutoniumPlutonium Registered User regular
    edited March 2010
    The probability of 1 appearing in the first 4 is 4/10.

    The probability of 2, 3, and 4 not being in the first four is 6/10 * 5/10 * 4/10 = 120/1000 = 12/100

    Which means the chances of the first four containing at least one of 2, 3, or 4 is 88/100

    Thus the probability of both having a 1 in the first four, and at least one of 2, 3, or 4 in the first four is 4/10 * 88/100 = 352/1000 = 35.2%

    So it's 35.2%




    Though I'm not sure if you're going for one or more of 2/3/4, or one but not more of them.

    Plutonium on
  • JohnDoeJohnDoe Registered User regular
    edited March 2010
    One or more.

    Edit - Hmm. Either my script is wrong or that isn't (exactly) correct. I'm getting roughly 30% of this happening.

    JohnDoe on
  • AegisAegis Fear My Dance Overshot Toronto, Landed in OttawaRegistered User regular
    edited March 2010
    I don't think that the probability of 2, 3, and 4 not being in the first four will be out of 10. 1 has to be in the top 4 according to the situation, and as such there should only be 9 other spots that 2, 3, or 4 can occur.

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  • HiroconHirocon Registered User regular
    edited March 2010
    The events

    (1 is in the first 4)

    and

    (2, 3, or 4 are in the first four)

    are not independent.

    The probability as I calculate it is

    (4/10) * [1 - (6/9)*(5/8)*(4/7)]

    (4/10) is the probability that 1 is in the first four. Given that 1 is in the first four, the probability that 2, 3 and 4 are all NOT in the first four is (6/9)*(5/8)*(4/7).

    EDIT: The answer reduces to 32/105, which is approximately 30.5%.

    Hirocon on
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