Question

10 coins are tossed what is the probability that exactly 5 heads appear?

Answer 1

Hint: We solve this question by first finding the total number of outcomes possible when 10 coins are tossed. Then we find the number of ways we can get 5 heads when 10 coins are tossed using the formula, the number of ways of arranging $m+n$ objects out of which $m$ are the same and $n$ are the same is $\dfrac{\left( m+n \right)!}{m!\times n!}$. Then we use the formula for probability, $P\left( A \right)=\dfrac{\text{number of outcomes of event A}}{\text{Total possible outcomes}}$ to find the required probability.

Complete step-by-step solution:
We are given that 10 coins are tossed.
When a coin is tossed there are only two possible outcomes either a head or a tail.
So, when 10 coins are tossed, each coin will have 2 possible outcomes. So, total number of outcomes when 10 coins are tossed are
$\Rightarrow 2\times 2\times 2\times 2\times 2\times .........\left( \text{10 times} \right)$
Now let us consider the formula
$a\times a\times a\times a\times a\times ........\left( \text{n times} \right)={{a}^{n}}$
So, we the total possible outcomes when 10 coins are tossed as,
$\Rightarrow 2\times 2\times 2\times 2\times 2\times .........\left( \text{10 times} \right)={{2}^{10}}=1024$
We need to find the probability of getting exactly 5 heads.
First, let us find the number of ways of getting 5 heads.
If there are exactly 5 heads when 10 coins are tossed, then the other 5 are tails.
So, the number of ways of getting 5 heads and 5 tails is equal to the number of ways of arranging 5 heads and 5 tails.
So, we need to find the number of ways we can arrange 5 heads and 5 tails.
Now, let us consider the formula for the number of ways of arranging $m+n$ objects out of which $m$ are the same and $n$ are the same
$\dfrac{\left( m+n \right)!}{m!\times n!}$
Using this formula, we get the number of ways of arranging 5 heads and 5 tails as
$\Rightarrow \dfrac{10!}{5!\times 5!}=\dfrac{6\times 7\times 8\times 9\times 10}{1\times 2\times 3\times 4\times 5}=252$
So, the number of ways of arranging 5 heads and 5 tails is 252.
So, the number of ways of getting exactly 5 heads when 10 coins are tossed is 252.
Now we need to find the probability of getting exactly 5 heads out of 10 tosses.
Let us consider the formula for probability of occurrence of an event A,
$P\left( A \right)=\dfrac{\text{number of outcomes of event A}}{\text{Total possible outcomes}}$
Using the above formula, we get
$P\left( \text{getting 5 Heads} \right)=\dfrac{\text{Number of ways of getting 5 heads}}{\text{Total number of outcomes}}$
So, now let us substitute the above obtained values in the formula. Then we get,
$\begin{align}
  & \Rightarrow P\left( \text{getting 5 Heads} \right)=\dfrac{252}{1024} \\
 & \Rightarrow P\left( \text{getting 5 Heads} \right)=\dfrac{63}{256} \\
\end{align}$
So, the probability of getting exactly 5 heads when 10 coins are tossed is $\dfrac{63}{256}$.
Hence answer is $\dfrac{63}{256}$.

Note: There is a possibility of making a mistake while finding the number of ways of arranging 5 heads and 5 tails. One might make a mistake by taking its value as $10!$ without considering the fact that all 5 heads are similar and all 5 tails are also similar. But it is wrong because as even if we interchange two heads the outcome is similar.