Question

If two coins are tossed 5 times, then what is the probability of getting 5 heads and 5 tails.
A. \[\dfrac{{63}}{{256}}\]
B. \[\dfrac{1}{{1024}}\]
C. \[\dfrac{2}{{205}}\]
D. \[\dfrac{9}{{64}}\]

Answer 1

Hint: There are two possible outcomes. Thus, we can apply the binomial distribution formula \[P\left( x \right) = {}^n{C_r}{\left( p \right)^r}{\left( q \right)^{n - r}}\].

Formula Used:
Binomial distribution: \[P\left( x \right) = {}^n{C_r}{\left( p \right)^r}{\left( q \right)^{n - r}}\]
\[p = \]probability of success
\[q = \]probability of failure.
\[{}^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}}\]

Complete step by step solution:
Total number of outcomes is \[5 + 5 = 10\].
The probability of getting a head after tossing a coin is \[ = \dfrac{1}{2}\]
The probability of getting a tail after tossing a coin is \[ = \dfrac{1}{2}\]
The probability of success is \[p = \dfrac{1}{2}\].
The probability of failure is \[q = \dfrac{1}{2}\].
Now applying binomial distribution \[P\left( x \right) = {}^n{C_r}{\left( p \right)^r}{\left( q \right)^{n - r}}\].
\[P\left( x \right) = {}^{10}{C_5}{\left( {\dfrac{1}{2}} \right)^5}{\left( {\dfrac{1}{2}} \right)^{10 - 5}}\]
Apply the formula \[{}^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}}\]
\[ = \dfrac{{10!}}{{\left( {10 - 5} \right)!5!}}{\left( {\dfrac{1}{2}} \right)^5}{\left( {\dfrac{1}{2}} \right)^5}\]
Apply the formula \[{a^m} \cdot {a^n} = {a^{m + n}}\]
\[ = \dfrac{{10!}}{{5!5!}}{\left( {\dfrac{1}{2}} \right)^{10}}\]
\[ = \dfrac{{10 \times 9 \times 8 \times 7 \times 6 \times 5!}}{{5!5!}}{\left( {\dfrac{1}{2}} \right)^{10}}\]
\[ = \dfrac{{10 \times 9 \times 8 \times 7 \times 6}}{{5!}} \times \dfrac{1}{{{2^{10}}}}\]
\[ = \dfrac{{30240}}{{120}} \times \dfrac{1}{{1024}}\]
\[ = \dfrac{{63}}{{256}}\]

Hence option A is the correct option.

Note: There are two possible outcomes after tossing a coin and each observation is independent that’s why we can apply the binomial theorem. In the given question there are two outcomes one is head and another is tail. Thus, we can apply binomial distribution.