Categories Conditional Probability and the Rules of Probability Probability What are your chances of flipping exactly 2 tails? Post author By mpershan Post date June 7, 2013 2 Comments on What are your chances of flipping exactly 2 tails? Can you explain where each of the student’s responses came from? As in, how did he arrive at 8/24, 14/96, and what is either 24/72 or 14/72. Share this:EmailPrint ← What are your chances of rolling a seven? → Odd Pairing 2 replies on “What are your chances of flipping exactly 2 tails?” I have no idea. My best guesses so far: Four coins with two sides each makes 8. Also the tree diagram at the right has 8 endpoints (though I’m not quite sure what the algorithm was for deciding when to end; it was sometimes after 4 flips OR after reaching two tails, but other times it stopped after 2 flips even when there were only 0 or 1 tails). There are 4 coins, and we want 2 tails, and 2*4 = 8. The list of H’s and T’s on the left has six possibilities listed, of which two have two Ts (marked by the student), so that suggests that the 1/3 (8/24) answer is reasonable. There are 2 sides per coin, 4 coins, thus by appending the digits we get 24. Or, more likely, there are 4 coins thus 4! = 24 ways to put them in order. There are 2 dice, and we want to get a 7, hence 2 * 7 = 14. Each die has 6 sides, and there are 2 dice, so 2 * 6 * 6 = 72. 14 / (24 + 72) = 14/96. I’m mostly pretty stumped, though — does anyone have some insight from the actual student in question? I agree with Josh for part a. I think the student was relying on their list. They made a list of six under 1 of the coins they drew. 2 of which had two tails. Therefore if they take the 6 possibilities and multiply it by four (thinking there are still 6 more lists to be made under each coin) they would get 24. Again, 2 of the possibilities met the requirement which means that 2×4 or 8 is how to get the numerator. I agree with Josh’s assumptions for part b, especially without the presence of any work shown. However if the student made a case that based on part a they just need multiply the parts to find the numerator (This could be just grab and use a known operation too). For part c, I think they kept the 14 because the chances of rolling a seven as they determined it was less likely than flipping two heads therefore that would be the one to use… I wonder how many students answered 0 to part b and c. I would have. I have no chance to roll a ‘7’ assuming I have two six-sided die. Technically you should reword it to say “What is the probability that the sum of two die when rolled together would equal 7″…just the assessment geek in me. Comments are closed.