What is the probability that he counts head more that tails when tosses a coin 6 times

$ \begingroup $ The first base footstep is to think of the unlike ways that it is potential to have Kitty count more heads ( H ) than tails ( T ). So the only way that that can occur is

1. Kitty gets 4 heads and 2 tails
2. Kitty gets 5 heads and 1 tail
3. Kitty
   gets 6 heads and 0 tails

now, any other scenario that occurs would not allow Kitty to have more heads than tails .
The next mistreat is then to calculate the probability of 1., 2., and 3. occurring.

naively, you could try to enumerate and count all of the different possibilities, but that would be a very long list to have to keep path of .
alternatively, let ‘s consider the follow : Let ‘s define some notation. Let $ X $ be the issue of heads that Kitty gets when tossing 6 coins. And so based on 1., 2., and 3. we are concern in the calculating the pursuit three probabilities :
$ Pr ( X=4 ) $, $ Pr ( X=5 ) $ and $ Pr ( X=6 ) $. Why are these relevant ? Think about it this way : We want to calculate
\begin { align* } Pr ( \text { Kitty gets more heads than tails } ) & =Pr ( X > 3 ) \\ & =Pr ( X=4 ) +Pr ( X=5 ) +Pr ( X=6 ) \\ & = { 6\choose 4 } ( 0.5 ) ^4 ( 0.5 ) ^2+ { 6\choose 5 } ( 0.5 ) ^5 ( 0.5 ) ^1+ { 6\choose 6 } ( 0.5 ) ^6 ( 0.5 ) ^0\\ & =15\times ( 0.5 ) ^6+6\times ( 0.5 ) ^6+1\times ( 0.5 ) ^6\\ & = ( 0.5 ) ^6\times ( 15+6+1 ) \\ & =0.34375 \end { align* }
The above solution makes manipulation of the fact that if a random variable $ adam $ follows the Binomial distribution with proportion $ p $ and act of trials $ n $, then the probability of seeing $ ten $ “ successes ” in $ normality $ trials is
$ $ Pr ( X=x ) = { n\choose ten } p^x ( 1-p ) ^ { n-x } $ $
@ guest11, I merely realized you said you have n’t learned the binomial convention which is what the above answer depends on. Let me try to explain it with out it.

Imagine the like setup as above. So let $ X $ be the count of heads that Kitty gets when tossing 6 coins .
Without using the binomial formula ( or distribution ) we can say that for Kitty to get more heads than tails we need to calculate the follow :
\begin { align* } Pr ( \text { Kitty gets more heads than tails } ) & =Pr ( X > 3 ) \\ & =Pr ( X=4 ) +Pr ( X=5 ) +Pr ( X=6 ) \\ \end { align* }
But so without using the binomial convention, how can we solve this ? Well lease ‘s try to calculate each probability one at a prison term .
so, what the probability that Kitty gets 6 heads, i.e., $ Pr ( X=6 ) $. Well that can only happen one means. Kitty has to get the stick to sequence of coin flips HHHHHH. And we know ( or assume ) that the coin is fair and sol the probability of getting a head on any one somersault is $ \frac { 1 } { 2 } $ or 0.5. But we besides know that flipping coins are independent events and so we can say that $ $ Pr ( X=6 ) =Pr ( \text { Kitty gets } HHHHHH ) =\frac { 1 } { 2 } \times\frac { 1 } { 2 } \times\frac { 1 } { 2 } \times\frac { 1 } { 2 } \times\frac { 1 } { 2 } \times\frac { 1 } { 2 } = ( 0.5 ) ^6 $ $
That ‘s the simple one. But how can we calculate the probability that Kitty gets 5 heads ? well, there are 6 ways for that to happen : HHHHHT, HHHHTH, HHHTHH, HHTHHH, HTHHHH, THHHHH. And so we can say that the probability that Kitty gets 5 heads is \begin { align* } Pr ( X=5 ) & =Pr ( \text { Kitty gets } HHHHHT \text { or } HHHHTH \text { or } HHHTHH\\ & \text { or } HHTHHH \text { or } HTHHHH \text { or } THHHHH ) \\ & =Pr ( \text { Kitty gets } HHHHHT ) + Pr ( \text { Kitty gets } HHHHTH ) + Pr ( \text { Kitty gets } HHHTHH ) \\ & +Pr ( \text { Kitty gets } HHTHHH ) +Pr ( \text { Kitty gets } HTHHHH ) + Pr ( \text { Kitty gets } THHHHH ) \\ & =\left ( \frac { 1 } { 2 } \times\frac { 1 } { 2 } \times\frac { 1 } { 2 } \times\frac { 1 } { 2 } \times\frac { 1 } { 2 } \times\frac { 1 } { 2 } \right ) + \left ( \frac { 1 } { 2 } \times\frac { 1 } { 2 } \times\frac { 1 } { 2 } \times\frac { 1 } { 2 } \times\frac { 1 } { 2 } \times\frac { 1 } { 2 } \right ) \\ & +\left ( \frac { 1 } { 2 } \times\frac { 1 } { 2 } \times\frac { 1 } { 2 } \times\frac { 1 } { 2 } \times\frac { 1 } { 2 } \times\frac { 1 } { 2 } \right ) + \left ( \frac { 1 } { 2 } \times\frac { 1 } { 2 } \times\frac { 1 } { 2 } \times\frac { 1 } { 2 } \times\frac { 1 } { 2 } \times\frac { 1 } { 2 } \right ) \\ & +\left ( \frac { 1 } { 2 } \times\frac { 1 } { 2 } \times\frac { 1 } { 2 } \times\frac { 1 } { 2 } \times\frac { 1 } { 2 } \times\frac { 1 } { 2 } \right ) +\left ( \frac { 1 } { 2 } \times\frac { 1 } { 2 } \times\frac { 1 } { 2 } \times\frac { 1 } { 2 } \times\frac { 1 } { 2 } \times\frac { 1 } { 2 } \right ) \\ & =6\times\left ( \frac { 1 } { 2 } \times\frac { 1 } { 2 } \times\frac { 1 } { 2 } \times\frac { 1 } { 2 } \times\frac { 1 } { 2 } \times\frac { 1 } { 2 } \right ) \\ & =6\times ( 0.5 ) ^6 \end { align* }
And so you see that we have

\begin { align* } Pr ( \text { Kitty gets more heads than tails } ) & =Pr ( X > 3 ) \\ & =Pr ( X=4 ) +Pr ( X=5 ) +Pr ( X=6 ) \\ & =Pr ( X=4 ) +6\times ( 0.5 ) ^6+6\times ( 0.5 ) ^6 \end { align* }
Which is matching our answer using the bionmial rule. We could similarly solve for $ Pr ( X=4 ) $ but in that case there are 15 ways to write down the phone number of ways of getting 4 heads and 2 tails ( but I think at this point you should be able to do it and convinced that you do n’t want to ! ) .
And so hopefully you see why the binomial formula is such a utilitarian creature .

source : https://leowiki.com
Category : Economy