anonymous
  • anonymous
If a fair coin is tossed four times, what is the probability of obtaining at most one head? I know the answer is 5/16 but how do you get that? Please Help!
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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phi
  • phi
The formula for n "successes" out of m tries is n Choose m p^n * q^(n-m) where p is the probability of success and q is 1-p (probability of failure) in your case 4C1 * (1/2)^1 * (1/2)^3 4 * 1/2* 1/8 = 4/16 at most one head means 1 head or 0 heads the chance for 0 heads is 4C0 * (1/2)^0 * (1/2)^4 1 * 1 * 1/16 1/16 so the chance of 1 head or 0 heads is 4/16 + 1/16 = 5/16
anonymous
  • anonymous
thank you- why are you multiplying 4C1 by (1/2)^1 and (1/2)^3
phi
  • phi
you could work it out in detail, by saying the chance of 0 heads is TTTT 1/2 chance of 0 heads on the first toss times 1/2 chance of heads on the 2nd toss, etc to get 1/2 * 1/2 * 1/2 * 1/2 = 1/16 chance of no heads the chance of 1 head has 4 cases: HTTT 1/2 chance of heads on the first toss times 1/2 chance of tails on the 2nd toss * 1/2 chance of tails on 3rd toss * 1/2 chance of tails on the 4th toss = 1/2 *1/2 * 1/2 * 1/2 = 1/6 plus THTT TTHT TTTH you can work out that each has a chance of 1/16 all together you get 4/16 chance of 1 head and 1/16 chance of 0 heads, or 5/16 chance of 0 or 1 head

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anonymous
  • anonymous
Oh I get it! thank you so much!
phi
  • phi
the 4C1 is the count of the ways of getting one head in 4 tries. The 4 combinations are showed up above. It gets more complicated for say, 3 heads in 5 tries. you can work out the combinations, but there will be 5C3 of them

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