A fair coin is tossed 5 times. What is the prob that it lands heads up at least twice?
you know there is a formula for this but i cannot remember it… so right now i’ll just do the “long way”, which means to think about this from the other side of the question. The probability that the coin will show AT LEAST two tails is (1 - the probability of the coin showing one or no heads) # of events when the coin shows NO heads = 1 # of events when the coin showed 1 head = 5 (think about it, either in the first toss, second toss, third toss, fourth toss or fifth toss) Total unfavourable events = 6 (stated above) Total events = 2^5 = 32 (heads or tails five times) Therefore, probability of showing AT LEAST two tails = 1 - 6/32 = 26/32 = 13/16 or 81.25%
2 to the fifth possible outcomes, or 32 possible combos. One way to have zero heads (all tails) Five ways to have exactly one heads So its (32-6)/32, or 81.25%
Easiest way to conceptualize this is to find the probability that heads comes up exactly zero times and add it to the probability that heads comes up exactly one time. You can add these because they are mutually exclusive. The sum you calculate will give you the complement to your question, so you have to subtract this total from 1 (100%). Give that a shot, and see if it helps you understand the material. If it’s still not clicking for you, feel free to write out each step you took and the calculations attempted, so that forum members can help find your mis-step.
ANyone know the formula? Why is the # of events when the coin shows NO heads = 1? i think the part of the AT LEAST two is really throwing me off…
Formula is nCr p^r (1 -p)^(n -r) So probability of at least 2 = 1 - probab of less than 2 (less calculations) p(<2) = p(0) + p(1) p(0) = 5C0 * .5^0 * .5^5 = .03125 p(1) = 5C1 * .5 * (.5^4) = …15625 p(<2) = .1875 p(at elast 2) = 1 - .1875 = .8125
CFA2010 Wrote: ------------------------------------------------------- > Why is the # of events when the coin shows NO > heads = 1? uhhh… if you don’t get this than the probability that you will ever fully grasp probaility is very low. You toss a coin. It’s either heads or tails. With 5 tosses you can look at (all possible)outcomes in this format : H or T… H or T … H or T … H or T … H or T Once again, there are 2^ 5, or 32 possible letter combos. There is only one way to have all tails T T T T T There are only 5 ways to have exactly one heads H T T T T T H T T T T T H T T T T T H T T T T T H > i think the part of the AT LEAST two is really > throwing me off… Anything else has at least two letter H’s in it. Your choices are no letter H, 1 letter H, two letter H’s, 3 letter H’s , etc. (sorry about the edits… would help if there was a built in spell check, or if I could type
CFA2010, If you want to go one step further, you can have a look at the definition of the Binomial distribution. Regards, Marc
as stated by mhannebert this question falls under binomial distribution…which there is a formula you could use. lets re word your example a bit. As opposed to choosing at least 2 heads lets just go with just 2. the formula is 5C2*(1/2)^2*(1/2)^3 this will give you .3125 now since your question states at least two then you have to add the probability of getting 2,3,4, and 5 heads which adds up to .8125 or you could subtract P(0), P(1) from 1.