Question

Let Z be a random variable giving the number of heads minus the number of tails in 5 tosses of a fair coin. Find the probability mass function of Z.

Answer 1

Answer:

haat bara... computer er question ey Z diye probaility r gaar marche .... bhagg

Answer 2

Answer:

The probability mass function is $\frac{1}{8}$

Explanation:

$\begin{array}{cc}\text { Sample points } & \omega \\ H H H & 3 \\ H H T & 1 \\ H T H & 1 \\ H T T & -1 \\ T H H & 1 \\ T H T & -1 \\ T T H & -1 \\ T T T & -3\end{array}$

The sample space  $W$ is

$S_{W}=\{-3,-1,1,3\}$

corresponding to $3 T, 1 H 2 T, 2 H 1 T$, and $3 H$ respectively.

To find the probability for each case:

$&P(0 \text { heads \& } 3 \text { tails })=\frac{3 !}{0 !(3-0) !}\left(\frac{1}{2}\right)^{0}\left(\frac{1}{2}\right)^{3}=\frac{1}{8}$

$&P(1 \text { head \& } 2 \text { tails })=\frac{3 !}{1 !(3-1) !}\left(\frac{1}{2}\right)^{1}\left(\frac{1}{2}\right)^{2}=\frac{3}{8}$

$&P(2 \text { heads \& } 1 \text { tail })=\frac{3 !}{2 !(3-2) !}\left(\frac{1}{2}\right)^{2}\left(\frac{1}{2}\right)^{1}=\frac{3}{8}$

$&P(3 \text { heads \& 0 tails })=\frac{3 !}{3 !(3-3) !}\left(\frac{1}{2}\right)^{3}\left(\frac{1}{2}\right)^{0}=\frac{1}{8}$

i.e.:

$\\P(W=-3)=\frac{1}{8} \\P(W=-1)=\frac{3}{8} \\P(W=1)=\frac{3}{8} \\P(W=3)=\frac{1}{8}\end{gathered}$