Getting Two Heads in Four Tosses of a CoinDate: 05/17/2000 at 22:01:23 From: Melissa Subject: Probability of two heads on four tosses Dear Dr. Math, The question I need to ask is: What is the probability of getting two heads on four flips of an unbiased coin? I have looked at your other answers, and think it would be 1/8 because: 1/2 + 1/2 + 1/2 + 1/2 = 1/8 Thanks so much for your help. Melissa Dismukes Date: 05/18/2000 at 13:42:43 From: Doctor TWE Subject: Re: Probability of two heads on four tosses Hi Melissa - thanks for writing to Dr. Math. I think what you meant was (1/2)*(1/2)*(1/2)*(1/2) = 1/16, but that would be the answer for getting 4 heads in 4 flips (or 0 heads in 4 flips). Let's draw a probability tree: /\ Toss: / \ / \ / \ T H 1st | | / \ / \ / \ / \ / \ / \ T H T H 2nd | | | | / \ / \ / \ / \ / \ / \ / \ / \ T H T H T H T H 3rd | | | | | | | | / \ / \ / \ / \ / \ / \ / \ / \ T H T H T H T H T H T H T H T H 4th | | | | | | | | | | | | | | | | 0H 1H 1H 2H 1H 2H 2H 3H 1H 2H 2H 3H 2H 3H 3H 4H 4T 3T 3T 2T 3T 2T 2T 1T 3T 2T 2T 1T 2T 1T 1T 0T * * * * * * Since it is an unbiased coin, each branch has a probability of 1/2, each outcome is equiprobable (P = 1/16), and the probability of tossing exactly 2 heads (outcomes marked with *) is 6/16 = 3/8. For a more general solution to this type of problem, search Dr. Math for "binomial probability" (without the quotation marks) using our archive search engine at: http://mathforum.org/mathgrepform.html We have many archived questions and answers on this topic. I hope this helps. If you have any more questions, write back. - Doctor TWE, The Math Forum http://mathforum.org/dr.math/ |
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