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I have 6 coins with probability for heads in toss as:

0.51 0.52 0.53 0.57 0.48 0.49

What is the probability of getting exactly 3 heads in 6 tosses? I can do it for a normal coin but a rigged coin with individual probability is messing me up!

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  • 2
    $\begingroup$ you just have to sum each possibility. $\endgroup$ – Don Thousand Apr 23 at 16:18
  • $\begingroup$ There are a lot of computations. I would write a computer program. $\endgroup$ – saulspatz Apr 23 at 16:56
  • $\begingroup$ Do you toss each coin once, or do you randomly select with replacement six times and toss the coin you get? $\endgroup$ – Ross Millikan Apr 23 at 18:43
  • $\begingroup$ A spreadsheet will be your best tool. $\endgroup$ – Graham Kemp Apr 24 at 4:33
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Assuming,

A=0.51, B=0.52, C=0.53, D=0.57, E=0.48, F=0.49,

the requried probability will be,

=ABC(1-D)(1-E)(1-F)+ABD(1-C)(1-E)(1-F)+ABE(1-C)(1-D)(1-F)+ABF(1-C)(1-D)(1-E)+ACD(1-B)(1-E)(1-F)+ACE(1-B)(1-D)(1-F)+ACF(1-B)(1-D)(1-F)+ADE(1-B)(1-C)(1-F)+ADF(1-B)(1-C)(1-E)+AEF(1-B)(1-C)(1-D)+BCD(1-A)(1-E)(1-F)+BCE(1-A)(1-F)+BCF(1-A)(1-D)(1-E)+BDE(1-A)(1-C)(1-F)+BDF(1-A)(1-C)(1-E)+BEF(1-A)(1-C)(1-D)+CDE(1-A)(1-B)(1-F)+CDF(1-A)(1-B)(1-E)+CEF(1-A)(1-B)(1-D)+DEF(1-A)(1-B)(1-C)

like that, lots of calculation, leaving the rest of the calculation for you :)

writing a program is better option

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  • $\begingroup$ @saulspatz writing a program is the best idea $\endgroup$ – aranyak Apr 23 at 16:58
  • $\begingroup$ sorry... you are right.... silly mistake... $\endgroup$ – aranyak Apr 23 at 18:17
  • $\begingroup$ thanks @kimchi_lover $\endgroup$ – aranyak Apr 23 at 18:28
  • $\begingroup$ how should u code something like this though? $\endgroup$ – bjoshi Apr 23 at 19:03

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