Q: "If a same coin is flipped 5 times, what is the probability of getting specifically 3 heads?"

A: we deserve to do this concern by illustration blanks strategy.for eg. $$left(dfrac12 ight)left(dfrac12 ight)left(dfrac12 ight)left(dfrac12 ight)left(dfrac12 ight)*10$$

but is there a method to execute this same problem using combinations?for eg. $$dfrac_5C_3_10C_5 = dfrac10252$$ so how should I gain $dfrac1032$ instead of $dfrac10252$?

I to be really not sure exactly how to perform this making use of combinations. Any aid would be greatly appreciated.

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Thanks

EDIT: sorry, i edited the question from mine previous inquiry modified question. This is the initial question. My bad.


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edited Apr 12 "19 at 13:26
Nick
request Apr 12 "19 in ~ 10:09
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First, a key element that the problem is that you have a collection of possible outcomes from the succession of coin flips, and each is equally likely.This is true only because you have a same coin, that is, that does no favor heads and does no favor tails.

It"s due to the fact that of the truth that you have actually a fair coin that you can even think about using a an approach where you placed the number of "desired" outcomes in the numerator of a portion and put the total variety of possible outcomes in the denominator.But since you have actually a same coin, you have the right to use that method.

Now let"s counting the variety of outcomes to put in the denominator.

A little notation will assist here. Let $H_n$ stand for the truth that the coin came up top on the $n$th flip. Allow $T_n$ stand for the fact that the coin came up tails on the $n$th flip.

You have noticed $10$ various things that might occur during the succession of flips:$$ H_1, T_1, H_2, T_2, H_3, T_3, H_4, T_4, H_5, T_5. $$That is, there are five flips, you might get a head on any of those 5 flips,and you might get a tail on any kind of of those 5 flips.

It is additionally true that among these $10$ points that could happen, in the finish there will be specifically $5$ of them that in reality did happen.You can think the this method there space $_10C_5$ possible outcomes for the succession of flips--that is, select $5$ that the $10$ points in the list.

But $_10C_5$ counts the number of ways to pick $5$ items native $10$ if it is possible to pick any 5 things from the list.Given the perform of $10$ things,$ H_1, T_1, H_2, T_2, H_3, T_3, H_4, T_4, H_5, T_5, $here is one collection of $5$ things we can select from that list:

$$ H_1, T_1, H_2, T_2, H_3. $$

What does this signify? It states the first flip came up heads but the very first flip likewise came up tails. Likewise for the second flip. The fourth and fifth flips did not produce any type of results at all. Coins carry out not job-related that way.

The idea that "choose $5$ the end of $10$" thus has you counting difficult outcomes as possible outcomes. That"s wrong. For this reason $_10C_5$ is merely not a correct means to count the possible variety of outcomes of five coin tosses.

The actual number of outcomes is $2^5 = 32.$This is a tiny enough number that you can actually create every possible sequence of coin tosses on a solitary piece of record in not too lot time. It appears this would be a beneficial exercise in ~ this allude in her studies. Make certain you write each possible sequence that tosses (e.g. $H_1T_2H_3H_4H_5$) exactly once, and also count them.It helps if you write the perform in a sequence with a pattern that helps you make certain you wrote every sequence and did no write any twice.

If you counting the number of outcomes the way, you acquire the result$$ frac_5C_32^5 = frac1032,$$which is the same thing you acquired with the "drawing blanks" method.Note that this way of figuring the probability does use combinationsin the one place where it really renders sense to usage combinations.

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By the way, the answer come out specifically the exact same whether you upper and lower reversal a fair coin $5$ times or draw $5$ times through replacement indigenous a bag through an equal variety of red and also blue balls. The number of balls in the bag has no impact on the an outcome as long as exactly half are red and half are blue.You room equally most likely to gain red or blue ~ above the first draw,and in one of two people of those cases, since you placed the ball earlier in the bag before drawing again (that"s what "with replacement" means), you again have equal number of red and blue balls prior to the second draw and you are equally likely to acquire red or blue.