A fair coin is tossed 12 times in how many ways can a coin land heads at least 10 times

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  How can I calculate the probability of getting at least 3 heads in 5 flips of a fair coin? So what is the probability of 3 or more heads?

  • "At least 3 heads" would be P(3) + P(4) + P(5) = 10/32 + 5/32 + 1/32 = 16/32 = .5 = 50%. i.e. there are 10 ways to get exactly 3, 5 ways to get exactly 4, and 1 way to get exactly 1; or a total of 16 ways out of a total of 32 possibilities.

• How can you get the same answer using the combination formula n!/k!(n-k)!?

  I am have been confused because he continues to switch between both methods randomly and I can't seem to get the formula to work.

  • 
    

    Okay. Here's the main formual you'll get in your math class:

    n! / r! (n-r)!

    n stands for the total number of items
    r stands for the number of items that are going to be used, organized, picked etc.

    So... If you have a class of 10 kids and only 3 people can earn 1st, 2nd, and 3rd place; then you're n value would be 10 and your r value would be 3.

    This would give you the equation: 10! / (10-3)!
    that equals: 10! / 7!
    You can then say 10 * 9 * 8 * 7! / 7!
    factor out the 7! and you're left with 10*9*8 which equal 720. :)

    That's just permutations.
    In combinations, the order of what you're choosing doesn't matter. For example, you can order any 3 of the 5 toppings for you pizza. It doesn't matter what order the toppings go on, as long as they go on. So, n would equal 5 and r would equal 3.
    However, you're formula will have changed.
    This time is will be:
    n! / r! (n-r)!
    So your problem will look like so:
    5! / 3! (5-3)!
    You reduce to 5! / 3! * 2!
    Factor out the 3! to get 5*4/ 2!
    20/2 = 10
    So you have 10 different arrangements of topping to choose from. :)
    It's very helpful to memorize the two different formulas, so here they are again:

    Permutations: n! / (n-r)! (here the order of your items does matter)
    Combinations: n! / r! (n-r)! (the order your items are put on/used does not matter)

• What if the coin is unfair (ex. it lands on heads 80% of the time)? How would we calculate the probability of getting 3 heads in 5 flips?

  • Then that would be (8/10)*(8/10)*(8/10)*(2/10)*(2/10)=0,02048 -- but that is for a specific case, where the first three of the flips land in heads and the rest in tails. now we need to add all the other possible arrengments. in this case the are 10 possible outcomes where we have 3 heads----(0,02048)*10=20.48%.

• what about order of tails? For example Ha Hb Ta Tb and Ha Hb Tb Ta are the same in this case just as Hb Ha Ta Tb and Ha Hb Ta Tb so why aren't we dividing different Tails combinations too?
  please help

  • When you multiply the different ways to get three heads, you do not double count the same permutation of Heads twice. In other words, you are not counting Ha Hb Ta Tb and then again Ha Hb Tb Ta. This does not happen in the multiplication of permutations of the Heads. Since the order of Ha Hb do not repeat more than once, then there is no other double count to be considered beside the change in order from Ha Hb to Hb Ha, which is double counting the fact of getting two Heads.

• What if instead of flipping a coin we were flipping a tetrahedron with faces numbered from 'A' to 'D'?
  Then what is the probability of getting three 'A's in 8 flips?

  Is it (0.25)^3 x (0.75)^5 x 8C3 = 20.76416015625%?

  • I don't think thats right, 0.085% of rolling three 'A's or 99.91% chance of not rolling three 'A's out of 8 rolls . I used mathematica to generate all 65,536 ways to arrange "A,B,C,D" for 8 spots or 4^8. I used java to go through all 65,536 ways and keep a tally for arrangements with exactly three 'A's. It found 13,608 arrangements out of 65,536 have exactly three 'A's, not 56. The probability of getting exactly three 'A's rolling a tetrahedron 8 times is 13608/65536 or 20.76%, not 0.085%, which makes sense.
    I think this only works for things like heads or tails, pass or fail, scored or missed, like 0 or 1.

• How can we relate 32 (number of all possibilities of 5 flips) to 10 (number of combinations of 3 in 5)? 32 isn't the number of combinations of anything (as far as I see it), right?

  • 10 is the number of ways in which we can get exactly 3 heads (or exactly 3 tails) out of the 5 flips. All of the results are:
    

    5 nCr 0 = 1
    5 nCr 1 = 5
    5 nCr 2 = 10
    5 nCr 3 = 10 <-- This is what the 10 is counting
    5 nCr 4 = 5
    5 nCr 5 = 1

    
    The 32 is counting the 1 + 5 + 10 + 10 + 5 + 1, which is the total number of possible outcomes when flipping a coin 5 times.

• If anyone can help me with this, that would be great.

  The probability of "A" passing an exam is 1/2 and for "B" is 1/3. The answer for both of them passing is 1/6 [1/2*1/3].

  If a dice is thrown twice, what is the probability of getting a one on both throw? The answer is 11/36 [(1/6)^2+((1/6)*(5/6)*2)].

  Now here is my problem. For the second question, why not (1/6)*(1/6) for the probability of getting a 1 on both dice as the prob. of both "A" and "B" happening in #Q 1 is multiplying both the prob. of their passing the test?

  Wow, I think that was a bit complicated and long. But if anyone can help, thanks. :)

  • I'm not sure where you got the 11/36 figure, but it isn't correct.

    Your logic is right. The probability of getting a 1 on both independent throws is (1/6)·(1/6)=1/36.

    Alternatively, you can think of the die throws as selecting from a 6x6 table at random, with each cell having an equal probability of being chosen. (Note that 2 and 3 is a separate cell from 3 and 2, etc.) Out of 36 cells, only one has the 1 and 1 option, so the probability is 1/36.

• Can someone explain the line at

  2:35

  :

  Ha can take 5 different buckets

  Usually when we have a deck of 52 cards, it is 52 ways of getting one card and so forth.

  Here why is not 3 ways to get a head?

  • Hi Charles,

    If we were rolling a 3-sided die, then that would be the case. But that is not what we are doing here.

    In this particular case, we have 5 open slots for 3 objects to go into.

    So, first we need to randonmly put in Object-1 into 1 of the 5 spots, giving us a 1/5 chance to put Object-1 in any of the 5 spots.
    Object-1 = (1/5)

    Object two now has 4 remaining spots to go into, so Object-2 has a 1/4 chance to go into any one of the spots.
    Object-2 = (1/4).

    Finally, following the same logic Object-3 has 3 remaining choices.
    Object-3 = (1/3).

    So, for any particular case we will have a (1/5 * 1/4 * 1/3) chance of getting that outcome. Or to simply find the total number of outcomes it would be 5 * 4 * 3. This is if are including cases of the same 3 spots being taken, but in different order.

    Make sense?

    Hope this helps,
    - Convenient Colleague

    Note: As you go farther you will learn combinatoric equations that people might tell you to memorize. Really, it is better just to think through the process I gave the example for, relative to your situation, and then take into account things like if we care about order. Thinking things logically is always better than hard-core memorization, which doesn't teach you anything.

• Three coins are tossed up in the air. What is the probably that two of them will land heads and one will land tails? This is from the Wonderlic test prep. The answer key has 3/8 as a solution. what is the process for working this out. I have tried to use the combinatorics and can not get the correct answer. thanks

  • First, flipping the three coins at the same time is the same as flipping them one at a time since the events are independent, so we can use the same process that Sal uses. With combinatorics, we take 3 flips and choose 2 heads, which is 3!/[(2!)(3-2)!] = 3*2*1/[(2*1)(1)] = 3. Note that this is the same number as taking 3 flips and choosing 1 tails ( 3!/[(1!)(3-1)!] ). Since the total number of outcomes is 2^3 = 8, the probability that three tossed coins results in 2 heads and 1 tails is 3/8.

• I really can't get the reason of using the combination formula for the result in the numerator (outcomes that satisfy our goal). My reasoning is stuck with the idea that identical combinations with "different" heads must be counted (even though I know this yields to a probability larger than 1 which is impossible) because I keep seeing them as different possible results. It's hard to explain ... but I tend to think visually and this is driving me crazy as I can't draw any idea in my mind that leads me to think correctly! Thanks in advance
  Stefano

  • You can think about it as if you were interested in to find the number of DIFFERENT ways I can give you 3 bananas (b) and 2 apples (a), giving you one fruit at a time. If I give you the fruits is this order: bbaab, you will hope a different way of delivering for the next time. Take a look at this new combination: bbaab, It doesn't have any difference with respect to the one indicated above, despite the fact I changed the order of bananas and apples.

    You need to get rid of these redundancies when you are dealing with situation where you cannot infer that one object is different from another or you can infer that the order doesn't matter. The same thing happens with the coins. Unless you were told to pick 5 coins from a bag where each coin is different from the remaining ones (to say, they're from different countries), maybe in that situation the order does matter (it depends on how the problem was stated too).

When a coin tossed 12 times?

When flipping a fair coin 10 times what is the probability of at least one head?

How likely is it that a fair coin will land on heads every time if it is tossed 10 times?

What is the expected number of heads that come up when a fair coin is flipped ten times?