A fair coin is thrown ten times. Find the probability that more heads arise than tails?Hi guys my question is a binomial theorem one I know that I'm just finding it hard to know what goes where. I get that in order to have more heads I must have 6 heads or more nut find it there to use the.

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A fair coin is thrown ten times. Find the probability that more heads arise than tails?Hi guys my question is a binomial theorem one I know that I&#039;m just finding it hard to know what goes where. I get that in order to have more heads I must have 6 heads or more nut find it there to use the.

Santiago Collier · Accepted Answer

Step 1Actually there are three possible outcomes, when tossing a coin 10 times, which are the following (and with their respective probabilities):1. More heads than tails: with probability p (this is the probability you are looking for).2. More tails than heads: with the same probability, i.e. p (due to symmetry!).3. Equal number of tails and heads: That is the probability of 5 heads, i.e.                              (                            10        5                              )                                (              1        2            )        5              (      1      −              1        2            )              10      −      5      Now, these three probabilities add up to 1 ...

Seamus Mcknight · Accepted Answer

Step 1This is a binomial setting as you&#039;ve mentioned. What we&#039;re basically looking to find is   P  (  H  &amp;gt;  5  ), or the probability of getting more than 5 heads. This is what &quot;More heads than tails&quot; means. As you know, when we do bionmcdf(n,p,k), or use the equation, it&#039;s more convenient to change this to a probability with 0 as our lower bound. Let&#039;s turn that   P  (  H  &amp;gt;  5  ) into:  P  (  H  &amp;gt;  5  )  =  1  −  P  (  H  ≤  5  )Cool. Let&#039;s look at the different variables we&#039;ll need for a binomial distribution.   n  =  number of trials,   p  =  probability of &quot;success&quot;,   k  =  number of &quot;successes&quot;. Guess what? We know everything!   n  =  10, because we&#039;re tossing 10 coins.   k  =  5, because we&#039;re looking for   P  (  H  ≤  5  ) so we want MAX 5 successes, and   p  =      1    2   because there&#039;s a 50% chance of getting a head.Step 2We can just plug it into the calculator, and we&#039;ll get:   1  −  b  i  n  o  m  c  d  f  (  10  ,  0.5  ,  5  )  =  37.6953  %. Equation-wise, recall that this is our equation:  P  (  X  ≤  k  )  =      ∑          k      =      0              k                                    (                            n        k                              )                          p    k    (  1  −  p      )          n      −      k      I am used to using the variable k twice, a little confusing but will do. For this problem, it will be:  P  (  H  &amp;gt;  5  )  =  1  −  P  (  H  ≤  5  )  =  1  −      ∑          k      =      0              5                                    (                            10        k                              )                                (              1        2            )        k              (              1        2            )              10      −      k      This will come out to be   ≈  37.6953  % as well.I did this one-sided because bioncdf(n,p,k) is one-sided, but you can do it like this too:  P  (  H  &amp;gt;  5  )  =      ∑          k      =      6              10                                    (                            10        k                              )                                (              1        2            )        k              (              1        2            )              10      −      k      I started from 6 because it doesn&#039;t include 5 (greater than).

A fair coin is thrown ten times. Find the probability that more heads arise than tails?

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