Question

The probability of getting $4$ heads in $8$ throws of a coin is

• A. $\frac{60}{256}$
• B. $\frac{70}{216}$
• C. $\frac{35}{128}$
• D. $\frac{27}{128}$

Answer 1

The correct option is C

$\frac{35}{128}$

Calculate the probability

Whenever, we throws a coin there are $2$ possible outcomes either head or tail

$\therefore$the no of outcomes $=2$

Now , if we throws a coin $8$ times

The total no of outcomes we get $Z=2^8=256$

Here, the number of favourable outcomes of getting $4$ heads are

$$\begin{gathered} {}^{8}C_4=\frac{8!}{4!\left(8-4\right)!} \\ \Rightarrow {}^{8}C_4=\frac{8\times 7\times 6\times 5\times 4\times 3\times 2\times 1}{4\times 3\times 2\times 1\times 4\times 3\times 2\times 1} \\ \Rightarrow {}^{8}C_4=70 \end{gathered}$$ $[$applying ${}^{n}C_r=\frac{n!}{r!\left(n-r\right)!}]$

Probability of getting $4$ heads

$$\begin{gathered} P\left(4\right)=\frac{no.offavorableoutcomes}{Totalno.ofoutcomes} \\ \Rightarrow P\left(4\right)=\frac{70}{256} \\ \Rightarrow P\left(4\right)=\frac{35}{128} \end{gathered}$$

Therefore, probability of getting $4$ heads in $8$ throws of a coin is $\frac{35}{128}$,so option (C) is the correct answer. ,6299eeab15c1250dc30e2b4f