What is the chance of either getting all heads or no heads on 4 tosses?
The first toss can be anything (H or T) The second toss must be the same as the first -> prob = 1/2 The third toss must be the same as the first -> prob = 1/2 The fourth toss must be the same as the first -> prob = 1/2 The prob of the 2nd, 3rd and 4th tosses being the same as the first = 1/2 * 1/2 * 1/2 = 1/8 = 0.125
Chance of getting heads in either of two tosses?
Here are all the possibilities: T T H T T H H H The changes to have head in either toss is 75% (0.75). You can also see it as the change to getting heads on either toss is equal to the chance on NOT getting both tails. You already calculated the chance of getting both tails (0.25). The probability of NOT having both tails is 1 - 0.25 = 0.75.
What is the chance of either getting all heads or no heads on 4 tosses while tossing a fair coin?
If you toss a coin once, there are two possiblities--head or tails. If you toss a coin twice, there are four possibilities-- heads/heads, heads/tails, tails/heads, tails/tails. With three tosses there are 8 possibilities. With 4 tosses there are 16 possibilities. IOW the number of possible outcomes is 2^t where t is the number of tosses. Each new toss doubles the number of possible outcomes. So with 4 tosses there are 16 possible outcomes. Of which ONE is all heads, and ONE is all tails. Either one has a probability of 1/16, or 0.0625.
IF I toss a coin 3 times, what is the probability of getting NO HEADS?!?
If it's a fair coin, then each time there's 1/2 chance each of heads and tails. Thus, the odds of any throw being a tail is 1/2. The probability of 3 throws being tails is 1/2*1/2*1/2=1/8. Now, for 2 heads there are 3 possibilities, ie the tail could be first, second or third. Any particular such possibility has a probability of 1/2*1/2*1/2=1/8. Since there are three such possibilities, it's a total of 3/8. Now add to the previous 1/8, and you have a total of 4/8 or 1/2. More broadly, there are 3 possibilities when you throw a coin 3 times. 0 heads, 1 head, 2 heads or 3 heads, which is the same as 3 tails, 2 tails, 1 tail, or 0 tail. Totally, these have a probability of 1. 0 heads should have the same probability as 0 tail (3 heads). 2 heads should have the same probablity as 2 tails (1 head). YOu can see that these cover all the options. Thus No heads or two heads are exactly half the probability ie 1/2.
Since the tosses are independent, we can just calculate the probability of obtaining one head and repeating that 3 times. Obviously, the probability of getting a head for one toss is [math]\frac{1}{2}[/math], our answer is [math]\frac{1}{2}\cdot \frac{1}{2}\cdot \frac{1}{2}=\boxed{\frac{1}{8}}[/math]This can also be shown by:Again, there are a total of 8 total possibilities, and only one has three heads, so the answer is [math]\frac{1}{8}[/math]
A fair coin tossed four times in succession.?
It's not always 1/2. The probability of tossing any particular combination decreases with the number of tosses you make. Think about tossing two heads in a row. While the probability of tossing heads or tails on any ONE throw is 1/2, the first throw has two possible outcomes, either heads or tails. In order to throw two heads in a row, you have to throw heads the first time, which only has a probability of 1/2. Then you have to again throw heads, which also has a probability of 1/2. The total probability of throwing heads twice in a row is then (1/2) x (1/2)= 1/4. You can follow the same logic through for any arbitrary number of tosses. While this works for four heads, it does not work for 3 heads and a tail, as there are multiple ways to arrive with 3 heads and one tail. You could throw tails first, then three heads; or heads heads tails heads, or some other combination with three heads and one tails. The easiest way to solve this problem is probably to just write down all the different possible outcomes of four consecutive tosses, and then figure out which fraction of all possible outcomes corresponds to one tails and three heads. The possibilities are: HHHH HHHT HHTH HHTT HTHH HTTH HTHT HTTT TTTT TTTH TTHT TTHH THTT THTH THHT THHH You have sixteen possible outcomes, and four of these correspond to three heads and one tails. So the probability is 4/16 = 1/4
The total number of ways four coins can be flipped is 2^4, which is 16, since there are four coins that can either land heads or tails. Now, with the condition that there are 3 heads comes that there is one tail, so the one tail can be the first, second, third, or fourth coin. This means that there are 4 ways to get one tail and 3 heads. This is a simplified way to find total combinations, but this can be extended to any number of coins with the combination formula, or (number of coins) choose (number of heads). So because you are interested in 4 outcomes out of the 16 possible outcomes, so the probability of getting 3 heads out of four flips is 4/16 or 1/4 or 0.25.
This is just Fibonacci. To get a sequence of 12 flips with no repeated heads, you can take any such sequence of 11 flips and add a tails on the end, or you can take any such sequence of ten flips and add tails, then heads. That covers all potential sequences. To make that a bit more formal:Call [math]S_k[/math] the number of sequences of [math]k[/math] tosses that don't have two consecutive heads.Call [math]S_{k,H}[/math] the number of sequences of [math]k[/math] tosses that don't have two consecutive heads and end in heads.Call [math]S_{k,T}[/math] the number of sequences of [math]k[/math] tosses that don't have two consecutive heads and end in tails.Then [math]S_k = S_{k,H} + S_{k,T}[/math] because every sequence ends in heads or tails.To end in heads, you must come from a sequence that is one shorter and ends in tails, so[math]S_{k,H} =S_{k-1,T}[/math]To end in tails, you just have to come from a sequence one shorter, so[math]S_{k,T} = S_{k-1}[/math]Adding these last two equations gives[math]S_k = S_{k-1,T} + S_{k-1} = S_{k-2} + S_{k-1}[/math]This is the famous Fibonacci recursion relation, with [math]S_1 =2[/math] and [math]S_2 = 3[/math]. Finally, there are [math]2^{12}[/math] total ways to flip 12 coins, each of which is equally likely. So the final answer to your question is [math]\frac{F_{14}}{2^{12}} =\frac{377}{4096}[/math]