Question

Suppose that a fair coin (probability of heads equals\(\frac{1}{2}\)) is tossed independently 10 times. Use the table of the binomial distribution given at the end of this book to find the probability that strictly more heads are obtained than tails.

Short answer

Probability that strictly more heads are obtained than tails is \(0.3759\)

Step-by-step solution

Given information

A fair coin is tossed 10 times.

Probability of head and tail is \(\frac{1}{2}\)

Computing the probability

X be no. of heads obtain.

Strictly more heads than tail obtain:

If\(X \in \left\{ {6,7,8,9,10} \right\}\)

Probability of this event is the number in the binomial table corresponding to

\(p = 0.5\)and\(n = 10\)for k=6,7,...10

By the symmetry of the binomial distribution:

\(\frac{{\left( {1 - {\rm P}\left( {X = 5} \right)} \right)}}{2}\)

\(\frac{{\left( {1 - 0.2461} \right)}}{2}\)

\( = 0.3759\)