Toss a Coin Six Times
Date: 02/07/98 at 16:59:43 From: Ruth Beldon Subject: Coin tossing probabilities A. Suppose a coin is tossed 6 times. What is the probability that 6 heads will occur? (Answer: 1/64) B. What is the probablity that 3 heads will occur? (Book answer: 5/16) 6/3 x 1/2 to 3rd power x 1/2 to 3rd power = 20x1/8x1/8 = 5/16 C. X = 2 6/2x 1/2 squared x 1/2 to 4th = 15x1/4x1/16 = 15/64 My question is: where did the 20 come from in part B and the 15 in part C? How was this answer arrived at? Thank you, R. Beldon
Date: 02/07/98 at 18:29:05 From: Doctor Mitteldorf Subject: Re: Coin tossing probabilities Dear Ruth, The way you calculate probabilities for n coin tosses is to count the different ways (different combinations) that the event you're looking at could happen. Say there are 6 tosses. The first toss can be either heads or tails. The second can be either heads or tails. 2*2 = 4. The third can be either heads or tails... so you end up with 2^6 = 64 possibilities. Only one of these has all heads. But there are more ways that you could get 3 heads. It could be the first, second, and third, or the first, second and fourth that are heads. Or maybe the first, second and fifth. Here's a complete list: 123,124,125,126,134,135,136,145,146,156,234,235, 236,245,246,256,345,346,356,456. That's 20 possibilities out of 64, or 20/64 = 5/16. The answer is related to Pascal's triangle. The 6th row is 1 6 15 20 15 6 1 The numbers add up to 64, and the middle one is 20. There is a formula for these numbers, which your book is referring to: rth number in nth row of Pascal Triangle (counting from zero): n! --------------- (n-r)! r! In your case, 6*5*4*3*2*1 in the numerator, 3*2*1 and 3*2*1 again in the denominator. -Doctor Mitteldorf, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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