Question

If you flip a coin \[10\] times what are the odds that you will get the same number of heads and tails ?

Answer 1

Hint: In the given question, we have to find the probability of the same number of heads or tails in \[10\] coin throws. This is the case of binomial probability distribution. We can solve it by using the formula \[P(x){ = ^n}{C_x}{p^x}{q^{n - x}}\] where getting heads can be taken as “success” and tails as “failure” or vice-versa.

Complete step by step answer:
The number of ‘Success' in a series of n experiments, where each time a yes-no question is posed, the Boolean-valued result is expressed either with success or yes or true or one (probability \[p\]) or failure/no/false/zero (probability \[q = 1 - p\]) in a binomial probability distribution.
The binomial distribution formula is for any random variable \[x\], given by;
\[P(x){ = ^n}{C_x}{p^x}{q^{n - x}}\]
Where, \[n = \] number of experiments, \[p = \] Probability of Success in a single experiment and \[q = 1 - p = \] Probability of Failure in a single experiment.

We can solve the sum as follows: the probability of getting same number of heads on tossing a coin-
\[p = \dfrac{1}{2}\]
So, we can say that-
\[q = 1 - p = 1 - \dfrac{1}{2} = \dfrac{1}{2}\]
Now the coin is tossed ten times so \[n = 10\]. Since we require the same number of heads and tails, it means that out of ten outcomes, five shall be heads and five shall be tails. Hence, we will get \[x = 5\]. Applying the formula, we get,
\[P(x = 5) = 10{C_5}{(\dfrac{1}{2})^5}{(\dfrac{1}{2})^{10 - 5}}\]
\[\Rightarrow P(x = 5) = 10{C_5}{(\dfrac{1}{2})^5}{(\dfrac{1}{2})^5}\]
On further simplification, then
\[P(x = 5) = (\dfrac{{10 \times 9 \times 8 \times 7 \times 6}}{{5 \times 4 \times 3 \times 2 \times 1}})(\dfrac{1}{{32}})(\dfrac{1}{{32}})\]
\[\Rightarrow P(x = 5) = \dfrac{{63}}{{256}}\]
\[\therefore P(x = 5) = 0.2461\]

Thus, the odds are \[0.2461\] i.e. \[24.61\% \].

Note: A single success/failure test is also called a Bernoulli trial or Bernoulli experiment, and a series of outcomes is called a Bernoulli process. For \[n = 1\], i.e. a single experiment, the binomial distribution is a Bernoulli distribution. The binomial distribution formula can also be written in the form of \[n\]-Bernoulli trials as: \[^n{C_x} = \dfrac{{n!}}{{x!(n - x)!}}\]. Hence, probability can be written as: \[P(x) = \dfrac{{n!}}{{x!(n - x)!}}{p^x}{q^{n - x}}\].