Solveeit Logo

Question

Question: If you flip a coin \[10\] times what are the odds that you will get the same number of heads and tai...

If you flip a coin 1010 times what are the odds that you will get the same number of heads and tails ?

Explanation

Solution

In the given question, we have to find the probability of the same number of heads or tails in 1010 coin throws. This is the case of binomial probability distribution. We can solve it by using the formula P(x)=nCxpxqnxP(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 pp) or failure/no/false/zero (probability q=1pq = 1 - p) in a binomial probability distribution.
The binomial distribution formula is for any random variable xx, given by;
P(x)=nCxpxqnxP(x){ = ^n}{C_x}{p^x}{q^{n - x}}
Where, n=n = number of experiments, p=p = Probability of Success in a single experiment and q=1p=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=12p = \dfrac{1}{2}
So, we can say that-
q=1p=112=12q = 1 - p = 1 - \dfrac{1}{2} = \dfrac{1}{2}
Now the coin is tossed ten times so n=10n = 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=5x = 5. Applying the formula, we get,
P(x=5)=10C5(12)5(12)105P(x = 5) = 10{C_5}{(\dfrac{1}{2})^5}{(\dfrac{1}{2})^{10 - 5}}
P(x=5)=10C5(12)5(12)5\Rightarrow P(x = 5) = 10{C_5}{(\dfrac{1}{2})^5}{(\dfrac{1}{2})^5}
On further simplification, then
P(x=5)=(10×9×8×7×65×4×3×2×1)(132)(132)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}})
P(x=5)=63256\Rightarrow P(x = 5) = \dfrac{{63}}{{256}}
P(x=5)=0.2461\therefore P(x = 5) = 0.2461

Thus, the odds are 0.24610.2461 i.e. 24.61%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=1n = 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 nn-Bernoulli trials as: nCx=n!x!(nx)!^n{C_x} = \dfrac{{n!}}{{x!(n - x)!}}. Hence, probability can be written as: P(x)=n!x!(nx)!pxqnxP(x) = \dfrac{{n!}}{{x!(n - x)!}}{p^x}{q^{n - x}}.