Question

Find mean of the probability distribution is given of the number of heads obtained in three flips of a balanced coin

Answer 1

Please see the attachment

Answer 2

Answer:

Mean of the probability distribution is given of the number of heads obtained in three flips of a balanced coin is $1.5$

Step-by-step explanation:

Let $\mathrm{X}$ denote the success of getting heads.

Therefore, the sample space is

$\mathrm{S}=\{\mathrm{HHH}, \mathrm{HHT}, \mathrm{HTH}, \mathrm{HTT}, \mathrm{THH}, \mathrm{THT}, \mathrm{TTH}, \mathrm{TTT}\}$

It can be seen that$\mathrm{X}$ can take the value of $0,1,2$ or 3 .

$\therefore \mathrm{P}(\mathrm{X}=0)=\mathrm{P}(\mathrm{TTT})$

$=\mathrm{P}(\mathrm{T}) \cdot \mathrm{P}(\mathrm{T}) \cdot \mathrm{P}(\mathrm{T})$

$=\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}$

$=\frac{1}{8}$

$\therefore \mathrm{P}(\mathrm{X}=1)=\mathrm{P}(\mathrm{HHT})+\mathrm{P}(\mathrm{HTH})+\mathrm{P}(\mathrm{THH})$

$=\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}+\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}+\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}$

$=\frac{3}{8}$

$&\therefore \mathrm{P}(\mathrm{X}=1)=\mathrm{P}(\mathrm{HHT})+\mathrm{P}(\mathrm{HTH})+\mathrm{P}(\mathrm{THH})$

$&=\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}+\frac{1}{2} \times \frac{1}{2}$

$&\therefore \mathrm{P}(\mathrm{X}=2)=\mathrm{P}(\mathrm{HHT})+\mathrm{P}(\mathrm{HTH})+\mathrm{P}(\mathrm{THH})$

$&=\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}+\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}+\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \\&=\frac{3}{8}$

$&\therefore \mathrm{P}(\mathrm{X}=3)=\mathrm{P}(\mathrm{HHH}) \\&=\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \\&=\frac{1}{8}$

Therefore, the required probability distribution is as follows.

Mean of $X E(X), \mu=\sum X_{i} P\left(X_{i}\right)$

$=0 \times \frac{1}{8}+1 \times \frac{3}{8}+2 \times \frac{3}{8}+3 \times \frac{1}{8}$

$=\frac{3}{8}+\frac{3}{4}+\frac{3}{8}$

$=\frac{3}{2}$

$=1.5$