Question

If you flip a coin 7 times, what is the probability of it being 4 tails and 3 heads?

-Can this be done without binomial?

Answer 1

you mean without \(\binom{7}{3}\)?

Answer 2

you have \(2^7=128\) elements in your sample space, so i would not want to count all the possibilities by hand.

Answer 3

Like, the formula nCx p^x q^(n-x) can be used but could you also do:
P( Four Tails) x P(3 heads) ?

Answer 4

answer is easy enough to get since 128 is your denominator and
\[\dbinom{7}{3}=\frac{7\times 6\times 5}{3\times 2}=35\]

Answer 5

you don't need all that since \(p=1-p=\frac{1}{2}\)

Answer 6

I'm pretty sure the binomial distribution would at least be the easiest way to do it.

Answer 7

and no, it is not the product

Answer 8

why is that?

Answer 9

you have 128 elements in your sample space all equally likely, so i would not even think of the formula that you wrote. it is just \(\frac{35}{128}\)

Answer 10

first off, i am not even sure what \(P(\text{4 tails})\) means. four tails in how many tries?

Answer 11

How did you get this?

Answer 12

saying" exactly 4 tails in 7 tries" is the same as saying "4 tails and 3 heads".

Answer 13

how did i get what?

Answer 14

Pascal's Triangle would give the 35 factor.

Answer 15

I'd like to interject once again saying that what what satellite is doing works very well in this case. However, if you have a biased coin, the binomial distribution is by far and away the best way to do it. (To my knowledge)

Answer 16

yes sorry i am trying to understand how satellite73 came to the anwer though. I cannot see it easily. I'm quite slow

Answer 17

yes of course. and you are uisng it in this case too, but it is not necessary to write \((\frac{1}{2})^4\times (1-\frac{1}{2})^3\) when it is simply \((\frac{1}{2})^7\)

Answer 18

@bobobobobb there are 35 possible cominations of 4 heads and 3 tails in 7 tosses.
since each outcome is equally likely, and there are 128 outcomes altogether, your answer is \(\frac{35}{128}\)

Answer 19

How did you get 35?

Answer 20

35=7C3

Answer 21

oh thank you @KingGeorge

Answer 22

and thanks @satellite73

Answer 23

yw
35 was computed here
\[\dbinom{7}{3}=\frac{7\times 6\times 5}{3\times 2}=35\]