What is the probability of tossing at least 1 head and 1 tail when you toss 6 coins?

What is the probability of tossing at least 1 head and 1 tail when you toss 6 coins?

Answers
First off: all heads HHHHHH
walterthewhale
21 September 2011
To answer this you need to consider ALL of the possible outcomes that you could get when you toss 6 coins. You need to treat each coin seperately & record all the possible results, ie, you could get all heads HHHHHH, or the last one could be a tail ie HHHHHT, or the penultimate one could be a tail, ie HHHHTH, or the last 2 HHHHTT, etc, With 6 coins there are a LOT of outcomes. In fact I make it 64 possible different results, try writing it out systematically, this is the hard bit, so I'll get you started
walterthewhale
21 September 2011
One tail HHHHHT, HHHHTH, HHHTHH, HHTHHH, HTHHHH, THHHHH, 6 ways of getting one tail.
walterthewhale
21 September 2011
Two tails: HHHHTT, HHHTHT, HHTHHT, HTHHHT, THHHHT, HHHTTH, HHTHTH, HTHHTH, THHHTH, HHTTHH, HTHTHH, THHTHH, HTTHHH, THTHHH, TTHHHH, so 15 ways of getting 2 tails, etc. When you have done them all there should be 64 total arrangements of the 6 coins. Now basic probabilty says that you look at the number of outcomes which satisfy your criteria ie, how many outcomes have at least one head & one tail. Luckily here this is failrly simple as all of the outcomes appart from all heads & all tails satisfy the criteria, so since 2 out of 64 do not meet the criteria, 62 out of 64 DO meet your criteria. So the answer is 62/64, which cancels to 31/32 or 96.875 % if you prefer percentages.
walterthewhale
21 September 2011
The question doesn't say whether you are thinking about the order the coins are tossed in. I took it that the order didn't matter. In which case there are just 7 possible outcomes: 6 heads, 5 heads and 1 tail, 4 heads and 2 tails, 3 heads and 3 tails, 2 heads and 4 tails, 1 head and 5 tails, 6 tails. And all the outcomes except the 'all heads' or 'all tails' contain at least 1 head and 1 tail, so the probability would be 5/7.
veronica
21 September 2011
Think about if all the coins were different types, eg, a 1p, 2p, 5p, 10p, 20p, 50p. Get them out look at them if it helps. Now put them all to heads. This is one arrangement (with all heads). Now turn over the 50p so that it is tails, this is one arrangement to have one tail. Now put the 50p back to heads & turn over the 20p to tails. This is another DIFFERENT arrangement where there is one tail, notice it is not the same as the other arrangement to make one tail. They must be counted seperately. There are 6 ways of making one tail (try it with your coins). Putting the coins in order is simply a way of keeping track of which arrangements you have already
walterthewhale
23 September 2011
Think about if all the coins were different types, eg, a 1p, 2p, 5p, 10p, 20p, 50p. Get them out look at them if it helps. Now put them all to heads. This is one arrangement (with all heads). Now turn over the 50p so that it is tails, this is one arrangement to have one tail. Now put the 50p back to heads & turn over the 20p to tails. This is another DIFFERENT arrangement where there is one tail, notice it is not the same as the other arrangement to make one tail. Each arrangement must be counted seperately. There are 6 ways of making one tail (try it with your coins). Now try turning over 2 coins to tails and notice how many different ways of doing that there are (15 ways as above). You MUST go through a rigorous proceedure to count the total number of arrangements with 6 coins.
walterthewhale
23 September 2011
Putting the coins in order is just a way of keeping track of how many arrangements you have for each possible outcome ie, 1 arrangement for all heads, 6 arrangements for one tail, 15 arrangements for 2 tails, etc. It is from the total number of arrangments that you make your probability calculations. ie, there are 64 total arrangements & 62 of them meet the criteria. But discussion is good, please test this for yourself, toss 6 coins & see how many times you get all heads or all tails, it will not be much more than 3.125% of the time (ie, pretty unlikely).
walterthewhale
23 September 2011
1:3
cookie
24 September 2011
Add an answer Cancel reply

Similar questions

Probability trees

Jim Sellers is thinking about producing a new type of electric razor for men. If the market were favorable he would get a return of 100,000, but if the market for this new type of razorwere unfavorable, he would lose60,000. Since Ron Bush is a good friend of Jim Sellers, Jim is considering the possibility of using Bush Marketing Research to gather additional information about the market for the razor. Ron has suggested that Jimeither use a survey or a pilot study to test the market. The survey would be a sophisticated questionnaire administered to a test market. It will cost 5,000. Another alternative is to run a pilot study. This would involve producing a limited number of the new razors and tying to sell them in two cities that are typical of American cities. The pilot study is more accurate e but is also more expensive. It will cost20,000. Ron Bush has suggested that it would be a good idea for Jim to conduct either the survey or the pilot before Jim makes the decision concerning whether to produce the new razor. ButJim is not sure if the value of the survey or the pilot is worththe cost. Jim estimates that the probability of a successful market without performing a survey or pilot study is0.5. Furthermore, the probability of a favorable survey result given a favorable market for razors is 0.7, and the probability of a favorable survey result given an unsuccessful market for razors is 0.2. In addition, the probability of an unfavorable pilot study given an unfavorable market is 0.9, and the probability of ann unsuccessful pilot study result given a favorable market for razors is 0.2.

Draw the decision tree for this problem without the probability values. Computer the revised probabilities needed to complete the decision, and place these values in the decision tree. What is the best decision for Jim? Use Expected MonetaryValue (EMV) as the decision criterion.