Ten Coin Flips, Four HeadsDate: 05/08/2001 at 18:21:06 From: Timi Subject: If you flip a coin 10 times, what is the probability of getting at least 4 heads? Hi! My professor and I disagree about solving this problem: If you flip a coin ten times, what is the probability of getting at least four heads? The way I see it is that there are 2^10 possible outcomes, since each separate toss has 2 outcomes: head or tail, and there are 10 tosses. So 10C4 (or C10,4?) / 2^10 = 20.5% The way my professor deals with this problem is: "Add the ninth row and then the tenth row, by adding the two numbers above in Pascal's triangle : 1 9 36 84 126 126 84 36 9 1 1 10 45 120 210 252 210 120 45 10 1 and we can write our equation: 1[(.5)^10] [(.5)^0] + 10[(.5)^9] [(.5)^1] + 45 [(.5)^8] [(.5)^2]+120[(.5)^7] [(.5)^3] +210[(.5)^6] [(.5)^4]+252[(.5)^5] [(.5)^5]+210[(.5)^4] [(.5)^6]+ etc. but I can stop here because it says AT LEAST 4 and that means 10,9,8..4 so I can stop. Now a shortcut. Because our probabilities are .5 heads and .5 tails, each term is [(.5)^.5] or [1/2^10] or 1/(2^10) or 1/1024 so I can just add the coefficients I need and divide by 1024. Adding 1 10 45 120 210 252 210 = 848 which we divide by 1024 = 82.8% " It seems a little bit complex to me, and our answers are different, too. Could you please let me know what you think about this problem? Thank you! Timi Date: 05/10/2001 at 14:43:57 From: Doctor Twe Subject: Re: If you flip a coin 10 times, what is the probability of getting at least 4 heads? Hi Timi - thanks for writing to Dr. Math. Your professor is right. Here is one way to look at the problem: Since the total of the probabilities of all outcomes must be 1, the probability of getting at least four heads is: p(>=4H) = 1 - [ p(0H) + p(1H) + p(2H) + p(3H) ] It's easier to add up these four probabilities and subtract from 1 than to add up: p(>=4H) = p(4H) + p(5H) + p(6H) + ... + p(10H) So what is the probability of getting no heads, p(0H) ? There's only one way to do that: T T T T T T T T T T so p(0H) = 1/1024. (Your assessment that there are 1024 possible combinations is correct.) What about the probability of getting exactly 1 head, p(1H)? There are ten ways to do that: H T T T T T T T T T T H T T T T T T T T T T H T T T T T T T : T T T T T T T T T H so p(1H) = 10/1024. What about the probability of getting exactly two heads, p(2H)? Here it starts to get tricky: H H T T T T T T T T H T H T T T T T T T H T T H T T T T T T That's 9 ways that start with H : H T T T T T T T T H T H H T T T T T T T T H T H T T T T T T T H T T H T T T T T That's 8 more ways that start with TH : T H T T T T T T T H T T H H T T T T T T T T H T H T T T T T T T H T T H T T T T That's 7 more ways that start with TTH : T T H T T T T T T H T T T H H T T T T T T T T H T H T T T T T T T H T T H T T T That's 6 more ways that start with TTTH : T T T H T T T T T H T T T T H H T T T T T T T T H T H T T T T T T T H T T H T T That's 5 more ways that start with TTTTH T T T T H T T T H T T T T T H T T T T H T T T T T H H T T T T T T T T H T H T T T T T T T H T T H T That's 4 more ways that start with TTTTTH T T T T T H T T T H T T T T T T H H T T T T T T T T H T H T That's 3 more ways that start with TTTTTTH T T T T T T H T T H T T T T T T T H H T T T T T T T T H T H That's 2 more ways that start with TTTTTTTH T T T T T T T T H H That's 1 more way starting with TTTTTTTTH So there are: 9 SUM i = 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 45 i=1 ways of getting exactly two heads, and so p(2H) = 45/1024. We can do the same for exactly 3 heads, getting: H H H T T T T T T T ... H H T T T T T T T H 8 combinations H T H H T T T T T T ... H T H T T T T T T H 7 combinations H T T H H T T T T T ... H T T H T T T T T H 6 combinations : : H T T T T T T T H H 1 combination T H H H T T T T T T ... T H H T T T T T T H 7 combinations T H T H H T T T T T ... T H T H T T T T T H 6 combinations T H T T H H H T T T ... T H T T H T T T T H 5 combinations : : T H T T T T T T H H 1 combination T T H H H T T T T T ... T T H H T T T T T H 6 combinations T T H T H H T T T T ... T T H T H T T T T H 5 combinations T T H T T H H H T T ... T T H T T H T T T H 4 combinations : : T T H T T T T T H H 1 combination and so on all the way to: T T T T T T T H H H 1 combination Thus we have: 8 7 6 5 4 3 2 1 SUM a + SUM b + SUM c + SUM d + SUM e + SUM f + SUM g + SUM h a=1 b=1 c=1 d=1 e=1 f=1 g=1 h=1 8 i = SUM [ SUM j ] i=1 j=1 = 1 + 3 + 6 + 10 + 15 + 21 + 28 + 36 = 120 ways of getting exactly three heads, so p(3H) = 120/1024, and our probability of getting at least four heads is: p(>=4H) = 1 - [1/1024 + 10/1024 + 45/1024 + 120/1024] = 1 - 176/1024 = 848/1024 = 53/64 ~= .828 (or 82.8%) Notice that the values in the numerators, i.e. 1, 10, 45 and 120, are the first four values in the 10th row of Pascal's triangle; a fact that your professor used to save the time needed to count the combinations. For more information on this, check out the following entries in our Ask Dr. Math archives: An Explanation of Pascal's Triangle http://mathforum.org/dr.math/problems/pascal.html Probability and Pascal's Triangle http://mathforum.org/dr.math/problems/ramirez6.4.98.html Toss a Coin Six Times http://mathforum.org/dr.math/problems/beldon2.7.98.html Binomial Expansions and Pascal's Triangle http://mathforum.org/dr.math/problems/fama.7.10.96.html I hope this helps! If you have any more questions or comments, write back again. - Doctor TWE, The Math Forum http://mathforum.org/dr.math/ |
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