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We can use binomial probability distribution in this case as follows:

We want number of heads as  $4$  in the first  $10$  tosses and number of tails as  $4$  in remaining  $10$  tosses.

Probability of getting  $4$ heads in the first  $10$  tosses can be calculated as :  $\binom{10}{4}\times\left ( \frac{1}{2} \right )^{4}\times\left ( \frac{1}{2} \right )^{6}$

Probability of getting $4$  tails in the remaining $10$  tosses can be calculated as : $\binom{10}{4}\times\left ( \frac{1}{2} \right )^{4}\times\left ( \frac{1}{2} \right )^{6}$ 

Since the probability of getting  $4$  heads the first  $10$  tosses is independent of getting  $4$  tails in the next  $10$  tosses, we get the answer by multiplying them as  :

$\rightarrow$ Probability of getting  $4$ heads in the first  $10$  tosses  \times  Probability of getting  $4$ tails in the

next  $10$  tosses   $=$   $\frac{210\times210}{2^{20}}= \frac{44100}{1048576}=0.042$  (when rounded off to 3 decimal places)

Refer: https://en.wikipedia.org/wiki/Binomial_distribution#Probability_mass_function

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