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Posted by on Friday, October 9, 2009 at 10:11pm.

This is a confusing problem to me also, can someone please help me?

What are the odds in favor of getting at least one head in three successive flips of a coin?

  • math - , Friday, October 9, 2009 at 10:16pm

    Pr(three heads)= 1/2*1/2*1/2
    Pr(one head in three flips)= Pr(HTT)+Pr(HHT) + Pr(HHH)+ Pr(THT)+Pr(TTH)+ Pr(THH)+Pr(HTH)
    = oddly enough 1/2*1/2*1/2 * 7
    check me to make certain I included all the ways to get at least one head.

    think on that one.

  • math - , Friday, October 9, 2009 at 10:16pm

    7/8

  • math - , Friday, October 9, 2009 at 10:18pm

    One comment. On the at least on head in three flips should equal the 1-Pr(no heads)0r 1=Pr(TTT) or 1-1/2*1/2*1/2= 7/8

  • math - , Friday, October 9, 2009 at 10:22pm

    post that same question in the little jiskha box search, and you'll find what you need.

  • math - , Friday, October 9, 2009 at 10:25pm

    Thanks alot!!!

  • math - , Friday, September 7, 2012 at 2:13am

    0.2DiVde dy1.5

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